Componendo-Dividendo and Value Putting

Competitive Mathematics · Algebra · SSC CGL

5 Part80 worked examples211 practice questions

Many algebra questions look long and frightening, yet their answer takes one or two lines. This chapter teaches the two methods behind that:

Componendo-Dividendo — when two fractions are equal, add and subtract the top and bottom of each to make new equal fractions. Questions with square roots, cube roots and fractions like $\frac{x + 1}{x - 1}$ open up with it.
Value Putting — replace $a, b, c, x$ by small, convenient numbers, turning the question into plain arithmetic, and knock out the wrong options.

Both methods carry one danger: forgetting the condition under which they work. So every method says when it works and when it cheats you. Read “When value putting fails” with special care: that is exactly how a correct-looking option turns out wrong in the exam.

If the question looks like	use this method
$\frac{a}{b} = \frac{c}{d}$ is given and something like $\frac{a + b}{a - b}$ is asked	Componendo-dividendo (methods 1, 2)
$x = \frac{A + B}{A - B}$	The square-root ladder (method 3)
$\frac{x ^{3} + 3 x}{3 x ^{2} + 1} =$ a number	The cube form (method 4)
$x = \frac{2 c d}{c + d}$ and $\frac{x + c}{x - c} + \frac{x + d}{x - d}$	Method 5
$\frac{a}{b} = \frac{c}{d} = \frac{e}{f}$	The $k$ method (method 8)
A sum of terms like $\frac{k}{x + k}$	The add-1 / subtract-1 move (method 10)
$a + b + c = 0$ , $pq + q r + r p = 0$ , $ab c = 1$	Condition questions (methods 11, 12) + value putting
Four options and an expression in letters	Value putting (from method 14)
“remainder on dividing by $x - 2$ ”, “value of $k$ ”	Remainder-factor shortcuts (method 18)
Least or greatest value	Method 20

Before you startThis chapter uses formulas such as

(a \pm b)^{2}

(a \pm b)^{3}

and

a^{3} + b^{3} + c^{3} - 3 ab c

. Their full treatment is in Algebraic Identities and the cleaning of square roots is in Indices and Surds; wherever needed the formula is given again here.

How it is laid outEvery method: idea → formula → how to spot → steps → solved examples → questions for you (try before you open the solution).

Componendo-Dividendo

Add, subtract, cancel

Making new equal fractions from equal fractions: the rules, the reverse use, square roots, cubes and special forms.

1. The three rules and why they hold

Componendo · dividendo · both

From $\frac{a}{b} = \frac{c}{d}$ to $\frac{a + b}{a - b} = \frac{c + d}{c - d}$ : adding, subtracting and cancelling.

The idea

Take two equal fractions $\frac{a}{b} = \frac{c}{d}$ , for instance $\frac{6}{4} = \frac{3}{2}$ . If you add the same number to equal things they stay equal. Add 1 on both sides: $\frac{a}{b} + 1 = \frac{c}{d} + 1$ , that is $\frac{a + b}{b} = \frac{c + d}{d}$ . This is componendo (“by adding”).

Now subtract 1 from both sides: $\frac{a - b}{b} = \frac{c - d}{d}$ , which is dividendo (“by subtracting”). Divide the first equality by the second: the bottoms $b$ and $d$ cancel and what remains is $\frac{a + b}{a - b} = \frac{c + d}{c - d}$ . This is componendo-dividendo, the form used most often.

◭ Diagram

Take

a : b = 5 : 3

a + b

holds 8 units and

a - b

holds 2, so

(a + b) : (a - b) = 8 : 2 = 4 : 1

. Any equal ratio such as

10 : 6

gives the same 4 : 1.

∑ Formula

\frac{a}{b} = \frac{c}{d} ⟹ \frac{a + b}{b} = \frac{c + d}{d} ⟹ \frac{a - b}{b} = \frac{c - d}{d} ⟹ \frac{a + b}{a - b} = \frac{c + d}{c - d}

The three rules

$a, b$: numerator and denominator of the first fraction
$c, d$: numerator and denominator of the second fraction

∑ Formula

\frac{a + b}{a - b} = \frac{c + d}{c - d} ⟹ \frac{a}{b} = \frac{c}{d}

The way back: the same rule once more

$a, b$: any two blocks (numbers or expressions)
$c, d$: the pair that gives their ratio

To remember: on top “first + second”, at the bottom “first − second”, on both sides. The way back is the same, so from $\frac{A + B}{A - B} = \frac{p}{q}$ you get $\frac{A}{B} = \frac{p + q}{p - q}$ at once. It is the $k$ method ( $A = p k$ , $B = q k$ ) written in one line.

Spot itUse it when the question has two equal fractions (or one fraction equal to a number such as

\frac{a}{b} = \frac{5}{1}

) and the fraction you want is built from the sum and difference of the given top and bottom.

Steps

Call the top of the given fraction “block $A$ ” and the bottom “block $B$ ” (a block may be an expression, like $3 x$ ).
Decide whether the asked fraction looks like $\frac{A + B}{A - B}$ or like $\frac{A}{B}$ .
Do the same move on both sides: plus on top, minus below.
Cancel common factors such as 2 and simplify.
Check with small numbers (for instance $A = 5 k, B = 3 k$ ).

Example 1 · Basic

\frac{x}{y} = \frac{7}{3}

, find the value of

\frac{x + y}{x - y}

Solution, step by step

1The blocks are $A = x$ and $B = y$ . Given $\frac{A}{B} = \frac{7}{3}$ , asked $\frac{A + B}{A - B}$ : exactly the form for componendo-dividendo.
2Plus on top, minus below: $\frac{x + y}{x - y} = \frac{7 + 3}{7 - 3} = \frac{10}{4} = \frac{5}{2}$ .
3The long way, as a check: take $x = 7 k, y = 3 k$ . Then $\frac{10 k}{4 k} = \frac{5}{2}$ , the same answer. The rule is only a short form of this $k$ calculation.

Answer $\frac{5}{2}$

If the asked fraction were upside down (

\frac{x - y}{x + y}

) the answer would be

\frac{2}{5}

; watch the order.

Example 2 · Standard

\frac{p + q}{p - q} = \frac{9}{5}

, find the value of

\frac{p}{q}

Solution, step by step

1The left side already has the form $\frac{A + B}{A - B}$ with $A = p$ , $B = q$ . We want $\frac{A}{B}$ , so we take the way back.
2Apply the rule once more: $\frac{( p + q ) + ( p - q )}{( p + q ) - ( p - q )} = \frac{9 + 5}{9 - 5}$ .
3The left side becomes $2 p$ over $2 q$ , so $\frac{2 p}{2 q} = \frac{14}{4}$ ⟹ $\frac{p}{q} = \frac{7}{2}$ .
4Check: $p = 7, q = 2$ gives $\frac{9}{5}$ ✓.

Answer $\frac{7}{2}$

Applying twice brings the original back; that is why the rule for “forward” and “back” is one and the same.

Example 3 · Standard

\frac{3 a + 2 b}{3 a - 2 b} = \frac{11}{3}

, find

\frac{a}{b}

Solution, step by step

1The blocks are $A = 3 a$ and $B = 2 b$ (whole terms, not just $a$ and $b$ ).
2The way back: $\frac{A}{B} = \frac{11 + 3}{11 - 3} = \frac{14}{8} = \frac{7}{4}$ , that is $\frac{3 a}{2 b} = \frac{7}{4}$ .
3Now remove the coefficients: $\frac{a}{b} = \frac{7}{4} \times \frac{2}{3} = \frac{14}{12} = \frac{7}{6}$ .

Answer $\frac{7}{6}$

The commonest slip is to take

\frac{3 a}{2 b} = \frac{7}{4}

itself as the answer for

\frac{a}{b}

Example 4 · Advanced

\frac{m ^{2} + n ^{2}}{m ^{2} - n ^{2}} = \frac{17}{8}

and

m, n

are positive, find

m : n

Solution, step by step

1The blocks are $A = m^{2}$ and $B = n^{2}$ , because the sum and the difference are of the squares.
2The way back: $\frac{m ^{2}}{n ^{2}} = \frac{17 + 8}{17 - 8} = \frac{25}{9}$ .
3Taking square roots, $\frac{m}{n} = \frac{5}{3}$ (both are positive), so $m : n = 5 : 3$ .

Answer $5 : 3$

When the blocks are squares you must take the root at the end; a negative solution such as

m = - 5 k

appears only if signs are free.

Example 5 · Standard

\frac{a - b}{b} = \frac{2}{5}

, find

\frac{a}{b}

Solution, step by step

1This is the dividendo shape ( $\frac{a - b}{b}$ ), so componendo undoes it: add 1 to both sides.
2 $\frac{a - b}{b} + 1 = \frac{2}{5} + 1$ ⟹ $\frac{a - b + b}{b} = \frac{7}{5}$ ⟹ $\frac{a}{b} = \frac{7}{5}$ .

Answer $\frac{7}{5}$

When only the denominator

b

is involved, “add or subtract 1” is the whole rule; you do not need full componendo-dividendo.

BasicTry it yourself first

\frac{a}{b} = \frac{9}{4}

, find the value of

\frac{a + b}{a - b}

BasicTry it yourself first

\frac{x}{y} = \frac{11}{5}

, find the value of

\frac{x + y}{x - y}

BasicTry it yourself first

\frac{m}{n} = \frac{8}{3}

, find the value of

\frac{m + n}{m - n}

BasicTry it yourself first

\frac{a + b}{a - b} = \frac{13}{3}

, find the value of

\frac{a}{b}

StandardTry it yourself first

\frac{x + y}{x - y} = \frac{7}{5}

, find the value of

\frac{x}{y}

StandardTry it yourself first

\frac{2 p + 3 q}{2 p - 3 q} = \frac{9}{5}

, find

\frac{p}{q}

StandardTry it yourself first

\frac{5 a + 2 b}{5 a - 2 b} = \frac{7}{3}

, find

\frac{a}{b}

StandardTry it yourself first

\frac{x - y}{x + y} = \frac{3}{7}

, find the value of

\frac{x}{y}

AdvancedTry it yourself first

\frac{a - b}{a + b} = \frac{2}{9}

, find the value of

\frac{a}{b}

AdvancedTry it yourself first

\frac{x ^{2} + y ^{2}}{x ^{2} - y ^{2}} = \frac{25}{7}

(

x, y

positive), find

\frac{x}{y}

AdvancedTry it yourself first

\frac{x ^{2} + y ^{2}}{x ^{2} - y ^{2}} = \frac{13}{5}

(

x, y

positive), find

\frac{x}{y}

2. Ratio from a condition, then the asked fraction

Two steps · match the blocks

Getting the ratio of two variables from the condition and then applying componendo-dividendo to the blocks of the asked fraction.

The idea

Often the question does not give a plain ratio; it gives an equation such as $\frac{2 x - y}{x + 3 y} = \frac{1}{2}$ and then asks the value of another fraction. There are two steps. Step 1: cross-multiply to get $x : y$ (an ordinary linear equation). Step 2: spot the blocks of the asked fraction, form the ratio of those blocks and apply componendo-dividendo.

The key point: componendo-dividendo works on the pair that is really on top and at the bottom of the asked fraction. If the asked fraction is $\frac{3 x + 4 y}{3 x - 4 y}$ the blocks are $3 x$ and $4 y$ , not $x$ and $y$ ; so first find $\frac{3 x}{4 y}$ , that is, bring the coefficients into $x : y$ .

Name	From $\frac{a}{b} = \frac{c}{d}$	Useful when
Invertendo	$\frac{b}{a} = \frac{d}{c}$	the fraction is given upside down
Alternendo	$\frac{a}{c} = \frac{b}{d}$	you need the ratio of the tops
Componendo	$\frac{a + b}{b} = \frac{c + d}{d}$	adding while keeping the bottom
Dividendo	$\frac{a - b}{b} = \frac{c - d}{d}$	subtracting while keeping the bottom
Componendo-dividendo	$\frac{a + b}{a - b} = \frac{c + d}{c - d}$	ratio of sum to difference

Here: $a, b$ are the top and bottom of the first fraction and $c, d$ of the second. Each rule turns equal fractions into equal fractions.

Spot itSpot it: the condition is an equation (not a ratio) and the asked fraction has different coefficients.

Steps

Cross-multiply and find $x : y$ .
Write the top and bottom blocks of the asked fraction.
Form the ratio of the blocks: multiply the terms of $x : y$ by their coefficients.
Apply componendo-dividendo and simplify.

Example 6 · Basic

\frac{4 x - y}{2 x + 3 y} = \frac{5}{6}

, find the value of

\frac{3 x + 4 y}{3 x - 4 y}

Solution, step by step

1Step 1, cross-multiply: $6 (4 x - y) = 5 (2 x + 3 y)$ ⟹ $24 x - 6 y = 10 x + 15 y$ ⟹ $14 x = 21 y$ , so $\frac{x}{y} = \frac{3}{2}$ .
2Step 2: the blocks of the asked fraction are $3 x$ and $4 y$ , so $\frac{3 x}{4 y} = \frac{3}{4} \times \frac{3}{2} = \frac{9}{8}$ .
3Componendo-dividendo: $\frac{3 x + 4 y}{3 x - 4 y} = \frac{9 + 8}{9 - 8} = 17$ .
4Check: at $x = 3, y = 2$ , $\frac{12 - 2}{6 + 6} = \frac{5}{6}$ ✓ and $\frac{9 + 8}{9 - 8} = 17$ ✓.

Answer $17$

Pushing

\frac{x}{y} = \frac{3}{2}

straight into the plus-minus step is wrong; let the coefficients of the blocks come in first.

Example 7 · Basic

x : y = 4 : 3

, find the value of

\frac{x ^{2} + y ^{2}}{x ^{2} - y ^{2}}

Solution, step by step

1The blocks are $x^{2}$ and $y^{2}$ , so first their ratio: $\frac{x ^{2}}{y ^{2}} = (\frac{4}{3})^{2} = \frac{16}{9}$ .
2Componendo-dividendo: $\frac{x ^{2} + y ^{2}}{x ^{2} - y ^{2}} = \frac{16 + 9}{16 - 9} = \frac{25}{7}$ .

Answer $\frac{25}{7}$

The ratio of square blocks is also squared; squaring

\frac{4 + 3}{4 - 3}

would be wrong.

Example 8 · Standard

\frac{y + v}{y - v} = \frac{9}{2}

and

\frac{x}{y} = \frac{u}{v}

, find the value of

\frac{x + u}{x - u}

Solution, step by step

1Given $\frac{x}{y} = \frac{u}{v}$ . By alternendo $\frac{x}{u} = \frac{y}{v}$ (multiply both sides by $\frac{y}{u}$ : the middle terms swap).
2Now componendo-dividendo on $\frac{x}{u} = \frac{y}{v}$ : $\frac{x + u}{x - u} = \frac{y + v}{y - v}$ .
3Hence $\frac{x + u}{x - u} = \frac{9}{2}$ .

Answer $\frac{9}{2}$

Without alternendo, componendo-dividendo gives

\frac{x + y}{x - y} = \frac{u + v}{u - v}

, a different (true, but useless here) relation.

Example 9 · Standard

x : y = 5 : 2

, find the value of

\frac{x + y}{x - y} + \frac{x - y}{x + y}

Solution, step by step

1Componendo-dividendo: $\frac{x + y}{x - y} = \frac{5 + 2}{5 - 2} = \frac{7}{3}$ .
2The second fraction is the reciprocal of this one, so it is $\frac{3}{7}$ ; no new calculation is needed.
3Sum $= \frac{7}{3} + \frac{3}{7} = \frac{49 + 9}{21} = \frac{58}{21}$ .

Answer $\frac{58}{21}$

BasicTry it yourself first

\frac{3 x - y}{x + 3 y} = \frac{7}{9}

, find the value of

\frac{2 x + y}{2 x - y}

BasicTry it yourself first

\frac{5 x - 2 y}{3 x + y} = \frac{8}{7}

, find the value of

\frac{2 x + 3 y}{2 x - 3 y}

StandardTry it yourself first

\frac{2 x - y}{x + y} = \frac{7}{8}

, find the value of

\frac{3 x + 2 y}{3 x - 2 y}

StandardTry it yourself first

\frac{3 x - 2 y}{2 x + y} = \frac{6}{11}

, find the value of

\frac{5 x + 3 y}{5 x - 3 y}

StandardTry it yourself first

\frac{4 x - 3 y}{x + 2 y} = \frac{16}{15}

, find the value of

\frac{2 x + 5 y}{2 x - 5 y}

AdvancedTry it yourself first

\frac{7 x - 2 y}{2 x + 3 y} = \frac{27}{22}

, find the value of

\frac{3 x + 4 y}{3 x - 4 y}

BasicTry it yourself first

x : y = 4 : 1

, find the value of

\frac{x + y}{x - y} + \frac{x - y}{x + y}

StandardTry it yourself first

x : y = 7 : 3

, find the value of

\frac{2 x + 3 y}{2 x - 3 y} + \frac{2 x - 3 y}{2 x + 3 y}

AdvancedTry it yourself first

x : y = 9 : 4

, find the value of

\frac{3 x + 5 y}{3 x - 5 y} + \frac{3 x - 5 y}{3 x + 5 y}

3. The fraction with square roots: the ladder

\frac{A + B}{A - B}

From $x = \frac{A + B}{A - B}$ : componendo-dividendo, square, componendo-dividendo again to get $\frac{x ^{2} + 1}{2 x} = \frac{A + B}{A - B}$ .

The idea

When $x = \frac{A + B}{A - B}$ , rationalising the denominator is long and risky. Componendo-dividendo untangles it in three steps like a ladder. Write $x$ as $\frac{x}{1}$ and apply the rule: adding the top and bottom gives $2 A$ , subtracting gives $2 B$ ; the 2 cancels and the roots stand apart cleanly.

◭ Diagram

The ladder: each step is one rule. The first step gives

A : B

, squaring gives

A : B

, and a second componendo-dividendo gives a neat relation in

x

∑ Formula

x = \frac{A + B}{A - B} ⟹ \frac{x + 1}{x - 1} = \frac{A}{B}, \frac{x ^{2} + 1}{2 x} = \frac{A + B}{A - B}

The square-root ladder

$A, B$: positive numbers or expressions, $A > B$
$x$: the given fraction

∑ Formula

x + \frac{1}{x} = \frac{2 ( A + B )}{A - B}

The same result in another dress

$x + \frac{1}{x}$: from which $x^{2} + \frac{1}{x ^{2}}$ , $x^{3} + \frac{1}{x ^{3}}$ and so on follow

Link with the $x + \frac{1}{x}$ chapter: $\frac{x ^{2} + 1}{x} = x + \frac{1}{x}$ , so $\frac{x ^{2} + 1}{2 x}$ is half of it. If the fraction is $x = \frac{A - B}{A + B}$ instead, $\frac{1 + x}{1 - x} = \frac{A}{B}$ and $x + \frac{1}{x}$ is the same again.

Spot itSpot it: the same two roots on top and bottom, one with

+

and the other with

-

; and the asked value is like

x^{2}

x + \frac{1}{x}

\frac{x ^{2} + 1}{x}

Steps

Identify $A$ and $B$ (the quantities under the roots).
First componendo-dividendo: $\frac{x + 1}{x - 1} = \frac{A}{B}$ .
Square both sides: $\frac{( x + 1 ) ^{2}}{( x - 1 ) ^{2}} = \frac{A}{B}$ .
Second componendo-dividendo: top $(x + 1)^{2} + (x - 1)^{2} = 2 x^{2} + 2$ , bottom $(x + 1)^{2} - (x - 1)^{2} = 4 x$ ⟹ $\frac{x ^{2} + 1}{2 x} = \frac{A + B}{A - B}$ .
Get whatever is asked (such as $x + \frac{1}{x}$ ) from this.

Example 10 · Basic

x = \frac{7 + 3}{7 - 3}

, find the value of

x + \frac{1}{x}

Solution, step by step

1Here $A = 7$ , $B = 3$ . The ladder gives $\frac{x ^{2} + 1}{2 x} = \frac{A + B}{A - B} = \frac{10}{4}$ .
2So $x + \frac{1}{x} = 2 \times \frac{10}{4} = 5$ .
3The long way (check): $x = \frac{( 7 + 3 ) ^{2}}{7 - 3} = \frac{10 + 2 21}{4} = \frac{5 + 21}{2}$ and $\frac{1}{x} = \frac{5 - 21}{2}$ (because $\frac{25 - 21}{4} = 1$ ). Their sum is $5$ ✓.

Answer $5$

The long way took three or four lines and a surd risk; the ladder took one line.

Example 11 · Standard

x = \frac{m + 2 n + m - 2 n}{m + 2 n - m - 2 n}

, find the value of

n x^{2} - m x + n

Solution, step by step

1Here $A = m + 2 n$ , $B = m - 2 n$ , so $A + B = 2 m$ and $A - B = 4 n$ .
2The ladder: $\frac{x ^{2} + 1}{2 x} = \frac{2 m}{4 n} = \frac{m}{2 n}$ .
3Cross-multiply: $2 n (x^{2} + 1) = m \cdot 2 x$ , that is $n (x^{2} + 1) = m x$ .
4Therefore $n x^{2} - m x + n = n (x^{2} + 1) - m x = 0$ .

Answer $0$

When a lettered question asks for a “simple form” it is usually a relation between

x

and the letters, and zero is a very common answer.

Example 12 · Standard

x = \frac{8 + 5}{8 - 5}

, find the value of

x^{2} + \frac{1}{x ^{2}}

Solution, step by step

1 $A = 8$ , $B = 5$ : $x + \frac{1}{x} = \frac{2 ( 8 + 5 )}{8 - 5} = \frac{26}{3}$ .
2By squaring, $x^{2} + \frac{1}{x ^{2}} = (x + \frac{1}{x})^{2} - 2$ (because $2 \cdot x \cdot \frac{1}{x} = 2$ ).
3 $= \frac{676}{9} - 2 = \frac{676 - 18}{9} = \frac{658}{9}$ .

Answer $\frac{658}{9}$

Example 13 · Advanced

x = \frac{13 + 12 + 13 - 12}{13 + 12 - 13 - 12}

, find

\frac{x ^{2} + 1}{2 x}

in two ways: by finding

x

directly, and by the ladder formula.

Solution, step by step

1Here $13 + 12 = 25$ and $13 - 12 = 1$ are perfect squares, so $x$ comes out directly: $x = \frac{5 + 1}{5 - 1} = \frac{3}{2}$ .
2Check with the ladder: $\frac{x ^{2} + 1}{2 x} = \frac{\frac{9}{4} + 1}{3} = \frac{13}{12}$ ✓.
3Lesson: when the roots are perfect squares the ladder and the direct calculation agree; this is how to test your own method.

Answer $\frac{13}{12}$

In the lettered form (

m

and

n

) the same number is

\frac{m}{n}

, here

\frac{13}{12}

BasicTry it yourself first

x = \frac{9 + 4}{9 - 4}

, find the value of

x + \frac{1}{x}

BasicTry it yourself first

x = \frac{11 + 5}{11 - 5}

, find the value of

x + \frac{1}{x}

StandardTry it yourself first

x = \frac{6 + 2}{6 - 2}

, find the value of

x^{2} + \frac{1}{x ^{2}}

StandardTry it yourself first

x = \frac{10 + 3}{10 - 3}

, find the value of

x^{2} + \frac{1}{x ^{2}}

StandardTry it yourself first

x = \frac{8 + 2}{8 - 2}

, find the value of

x - \frac{1}{x}

AdvancedTry it yourself first

x = \frac{18 + 8}{18 - 8}

, find the value of

x - \frac{1}{x}

AdvancedTry it yourself first

x = \frac{7 + 5}{7 - 5}

, find the value of

x^{3} + \frac{1}{x ^{3}}

AdvancedTry it yourself first

x = \frac{12 + 3}{12 - 3}

, find the value of

x - \frac{1}{x}

4. The cube form

\frac{x ^{3} + 3 x}{3 x ^{2} + 1}

Componendo-dividendo on $(x + 1)^{3}$ and $(x - 1)^{3}$ : $\frac{x ^{3} + 3 x}{3 x ^{2} + 1} = \frac{A + B}{A - B}$ .

The idea

When a cube replaces the square, the same ladder runs, but the second componendo-dividendo needs the expansion of $(x + 1)^{3}$ and $(x - 1)^{3}$ . Learn it once and reuse the result every time. $(x + 1)^{3} = x^{3} + 3 x^{2} + 3 x + 1$ and $(x - 1)^{3} = x^{3} - 3 x^{2} + 3 x - 1$ . Their sum is $2 x^{3} + 6 x = 2 (x^{3} + 3 x)$ (the $3 x^{2}$ and the $1$ cancel and the others double) and their difference is $6 x^{2} + 2 = 2 (3 x^{2} + 1)$ .

∑ Formula

(x + 1)^{3} (x - 1)^{3} = x^{3} + 3 x^{2} + 3 x + 1 = x^{3} - 3 x^{2} + 3 x - 1

The expansions needed

$x$: any number or expression

∑ Formula

x = \frac{3 A + 3 B}{3 A - 3 B} ⟹ \frac{( x + 1 ) ^{3}}{( x - 1 ) ^{3}} = \frac{A}{B}, \frac{x ^{3} + 3 x}{3 x ^{2} + 1} = \frac{A + B}{A - B}

The cube ladder

$A, B$: the numbers whose cube roots appear
$x$: the given fraction

It runs both ways. Backwards: if $\frac{x ^{3} + 3 x}{3 x ^{2} + 1} = \frac{p}{q}$ is given, componendo-dividendo gives $\frac{( x + 1 ) ^{3}}{( x - 1 ) ^{3}} = \frac{p + q}{p - q}$ . If this is the cube of a fraction, $(\frac{r}{s})^{3}$ , then $\frac{x + 1}{x - 1} = \frac{r}{s}$ and one more componendo-dividendo gives $x = \frac{r + s}{r - s}$ .

Spot itSpot it: the given fraction has

x^{3} + 3 x

(only

x^{3}

and

3 x

) on top and

3 x^{2} + 1

below; or

x

contains cube roots.

Steps

See the top and bottom as halves of $(x + 1)^{3} \pm (x - 1)^{3}$ .
Apply componendo-dividendo (form $\frac{p + q}{p - q}$ ).
Take the cube root to get $\frac{x + 1}{x - 1}$ .
One more componendo-dividendo: $x = \frac{r + s}{r - s}$ .

Example 14 · Basic

Simplify

(x + 1)^{3} + (x - 1)^{3}

and

(x + 1)^{3} - (x - 1)^{3}

Solution, step by step

1Expand both cubes: $(x^{3} + 3 x^{2} + 3 x + 1)$ and $(x^{3} - 3 x^{2} + 3 x - 1)$ .
2Adding keeps two $x^{3}$ and two $3 x$ terms and cancels $\pm 3 x^{2}$ , $\pm 1$ : sum $= 2 x^{3} + 6 x$ .
3Subtracting does the opposite: $x^{3}$ and $3 x$ cancel and $6 x^{2} + 2$ remains.

Answer $2 x^{3} + 6 x, 6 x^{2} + 2$

This is where the top

x^{3} + 3 x

and bottom

3 x^{2} + 1

come from (the 2 cancelled).

Example 15 · Standard

\frac{x ^{3} + 3 x}{3 x ^{2} + 1} = \frac{35}{19}

, find

x

Solution, step by step

1Componendo-dividendo: $\frac{( x + 1 ) ^{3}}{( x - 1 ) ^{3}} = \frac{35 + 19}{35 - 19} = \frac{54}{16} = \frac{27}{8}$ .
2 $27 = 3^{3}$ and $8 = 2^{3}$ , so taking cube roots, $\frac{x + 1}{x - 1} = \frac{3}{2}$ .
3The way back: $\frac{x}{1} = \frac{3 + 2}{3 - 2} = 5$ , that is $x = 5$ .
4Check: $\frac{125 + 15}{75 + 1} = \frac{140}{76} = \frac{35}{19}$ ✓.

Answer $x = 5$

This is a cubic equation; without componendo-dividendo it is hard to solve.

Example 16 · Standard

x = \frac{3 14 + 3 12}{3 14 - 3 12}

, find the value of

\frac{x ^{3} + 3 x}{3 x ^{2} + 1}

Solution, step by step

1Here $A = 14$ , $B = 12$ . First componendo-dividendo: $\frac{x + 1}{x - 1} = \frac{3 14}{3 12}$ .
2Cubing gives $\frac{( x + 1 ) ^{3}}{( x - 1 ) ^{3}} = \frac{14}{12}$ ; second componendo-dividendo: $\frac{x ^{3} + 3 x}{3 x ^{2} + 1} = \frac{14 + 12}{14 - 12} = \frac{26}{2} = 13$ .

Answer $13$

The ugly cube roots (

314

and so on) never had to be evaluated.

Example 17 · Advanced

x = \frac{3 p + q + 3 p - q}{3 p + q - 3 p - q}

, find the value of

q x^{3} - 3 p x^{2} + 3 q x - p

Solution, step by step

1 $A = p + q$ , $B = p - q$ , so $A + B = 2 p$ and $A - B = 2 q$ . The cube ladder: $\frac{x ^{3} + 3 x}{3 x ^{2} + 1} = \frac{2 p}{2 q} = \frac{p}{q}$ .
2Cross-multiply: $q (x^{3} + 3 x) = p (3 x^{2} + 1)$ , that is $q x^{3} + 3 q x = 3 p x^{2} + p$ .
3Bring everything to one side: $q x^{3} - 3 p x^{2} + 3 q x - p = 0$ .

Answer $0$

A special case, with

q = 1

, is

x^{3} - 3 p x^{2} + 3 x - p = 0

BasicTry it yourself first

x = \frac{3 9 + 3 7}{3 9 - 3 7}

, find the value of

\frac{x ^{3} + 3 x}{3 x ^{2} + 1}

BasicTry it yourself first

x = \frac{3 15 + 3 13}{3 15 - 3 13}

, find the value of

\frac{x ^{3} + 3 x}{3 x ^{2} + 1}

StandardTry it yourself first

x = \frac{3 20 + 3 12}{3 20 - 3 12}

, find the value of

\frac{x ^{3} + 3 x}{3 x ^{2} + 1}

StandardTry it yourself first

x = \frac{3 19 + 3 17}{3 19 - 3 17}

, find the value of

\frac{3 x ^{2} + 1}{x ^{3} + 3 x}

StandardTry it yourself first

\frac{x ^{3} + 3 x}{3 x ^{2} + 1} = \frac{9}{7}

, find

x

StandardTry it yourself first

\frac{x ^{3} + 3 x}{3 x ^{2} + 1} = \frac{35}{19}

, find

x

AdvancedTry it yourself first

\frac{x ^{3} + 3 x}{3 x ^{2} + 1} = \frac{91}{37}

, find

x

AdvancedTry it yourself first

\frac{x ^{3} + 3 x}{3 x ^{2} + 1} = \frac{341}{91}

, find

x

AdvancedTry it yourself first

\frac{x ^{3} + 3 x}{3 x ^{2} + 1} = \frac{76}{49}

, find

x

5. The $x = \frac{2 c d}{c + d}$ form

\frac{x + c}{x - c} + \frac{x + d}{x - d} = 2

When $x$ is the harmonic mean of two constants, the sum of two such fractions is always 2.

The idea

A favourite exam form: $x = \frac{4 ab}{a + b}$ (or $\frac{6 ab}{a + b}$ , or with roots, $\frac{4 15}{5 + 3}$ ) is given and $\frac{x + 2 a}{x - 2 a} + \frac{x + 2 b}{x - 2 b}$ is asked. You need not open $x$ at all. $\frac{2 c d}{c + d}$ is the harmonic mean of two constants $c, d$ , and these $c, d$ are exactly the numbers that sit beside $x$ in the fractions.

∑ Formula

x = \frac{2 c d}{c + d} ⟹ \frac{x + c}{x - c} + \frac{x + d}{x - d} = 2

The harmonic-mean result

$x$: the given value
$c, d$: the two constants that appear as $x \pm$ in the fractions

Why? Divide $x = \frac{2 c d}{c + d}$ by $c$ : $\frac{x}{c} = \frac{2 d}{c + d}$ . Componendo-dividendo gives $\frac{x + c}{x - c} = \frac{2 d + ( c + d )}{2 d - ( c + d )} = \frac{c + 3 d}{d - c}$ . Likewise dividing by $d$ gives $\frac{x + d}{x - d} = \frac{3 c + d}{c - d} = - \frac{3 c + d}{d - c}$ . Adding, $\frac{( c + 3 d ) - ( 3 c + d )}{d - c} = \frac{2 ( d - c )}{d - c} = 2$ .

The art of matching: $c$ and $d$ are the numbers that show in the fractions, so $x$ must be rewritable as $\frac{2 \times ( one ) \times ( other )}{one + other}$ . Example: $x = \frac{4 ab}{a + b}$ and the fractions have $x \pm 2 a$ , $x \pm 2 b$ ; take $c = 2 a$ , $d = 2 b$ : $\frac{2 ( 2 a ) ( 2 b )}{2 a + 2 b} = \frac{4 ab}{a + b}$ ✓. If the fractions had $x \pm a$ , $x \pm b$ then $x = \frac{4 ab}{a + b}$ would not be the harmonic mean and the result would not be 2.

Spot itSpot it:

x

has the shape “twice the product over the sum”, and the fractions carry

x +

and

x -

with those same two numbers.

Steps

Read $c$ and $d$ from the fractions.
Check that $x = \frac{2 c d}{c + d}$ (pull out coefficients and roots to see it).
If it does, the answer is $2$ . If not, apply componendo-dividendo to $\frac{x}{c}$ and $\frac{x}{d}$ separately.

Example 18 · Basic

x = 4

, find the value of

\frac{x + 6}{x - 6} + \frac{x + 3}{x - 3}

Solution, step by step

1From the fractions $c = 6$ and $d = 3$ . The harmonic mean is $\frac{2 \times 6 \times 3}{6 + 3} = \frac{36}{9} = 4$ , which is $x$ .
2So the sum is $2$ . Direct check: $\frac{4 + 6}{4 - 6} = - 5$ and $\frac{4 + 3}{4 - 3} = 7$ , sum $= 2$ ✓.

Answer $2$

One fraction is negative; do not worry, the sum is still 2.

Example 19 · Standard

x = \frac{6 ab}{a + b}

, find the value of

\frac{x + 3 a}{x - 3 a} + \frac{x + 3 b}{x - 3 b}

Solution, step by step

1In the fractions $c = 3 a$ and $d = 3 b$ . Check: $\frac{2 c d}{c + d} = \frac{2 \cdot 3 a \cdot 3 b}{3 a + 3 b} = \frac{18 ab}{3 ( a + b )} = \frac{6 ab}{a + b} = x$ ✓.
2 $\frac{x}{3 a} = \frac{2 b}{a + b}$ , so $\frac{x + 3 a}{x - 3 a} = \frac{2 b + a + b}{2 b - a - b} = \frac{3 b + a}{b - a}$ .
3 $\frac{x}{3 b} = \frac{2 a}{a + b}$ , so $\frac{x + 3 b}{x - 3 b} = \frac{2 a + a + b}{2 a - a - b} = \frac{3 a + b}{a - b} = - \frac{3 a + b}{b - a}$ .
4Adding: $\frac{( 3 b + a ) - ( 3 a + b )}{b - a} = \frac{2 ( b - a )}{b - a} = 2$ .

Answer $2$

Write these four lines for a full solution; otherwise spot the harmonic mean and write 2 at once.

Example 20 · Standard

x = \frac{60}{18 + 50}

, find the value of

\frac{x + 18}{x - 18} + \frac{x + 50}{x - 50}

Solution, step by step

1From the fractions $c = 18$ , $d = 50$ . Their product is $c d = 18 \times 50 = 900 = 30$ , so $2 c d = 60$ .
2Hence $x = \frac{2 c d}{c + d} = \frac{60}{18 + 50}$ , which is the given $x$ .
3Therefore the sum is $2$ .

Answer $2$

The numerator does not show

2 c d

directly; you must get

c d

from the product of the roots.

Example 21 · Advanced

Find

x

\frac{x + 3}{x - 3} + \frac{x + 6}{x - 6} = 2

Solution, step by step

1This is the form with $c = 3$ , $d = 6$ . The harmonic mean makes the sum exactly 2.
2 $x = \frac{2 \times 3 \times 6}{3 + 6} = \frac{36}{9} = 4$ .
3Check: $\frac{7}{1} + \frac{10}{- 2} = 7 - 5 = 2$ ✓. (No second value of $x$ exists: after clearing the fractions the equation is linear in $x$ .)

Answer $x = 4$

Warning:

x = 3

x = 6

is forbidden; the denominator would become zero.

BasicTry it yourself first

x = 8

, find the value of

\frac{x + 5}{x - 5} + \frac{x + 20}{x - 20}

BasicTry it yourself first

x = 6

, find the value of

\frac{x + 4}{x - 4} + \frac{x + 12}{x - 12}

StandardTry it yourself first

x = 12

, find the value of

\frac{x + 10}{x - 10} + \frac{x + 15}{x - 15}

StandardTry it yourself first

x = 12

, find the value of

\frac{x + 8}{x - 8}

StandardTry it yourself first

x = 8

, find the value of

\frac{x + 12}{x - 12}

AdvancedTry it yourself first

\frac{x + 10}{x - 10} + \frac{x + 40}{x - 40} = 2

, find

x

AdvancedTry it yourself first

\frac{x + 20}{x - 20} + \frac{x + 5}{x - 5} = 2

, find

x

AdvancedTry it yourself first

x = 20

, find the value of

\frac{x + 15}{x - 15} + \frac{x + 30}{x - 30}

6. When $ab$ , $2 ab$ or $a^{2} + b^{2}$ shows up

Make a perfect square

Questions with $(a + b) : ab$ , $\frac{a ^{2} + b ^{2}}{2 ab}$ and $x = \frac{2 am}{m ^{2} + 1}$ : componendo-dividendo produces perfect squares.

The idea

Two identities: $a + b \pm 2 ab = (a \pm b)^{2}$ and $a^{2} + b^{2} \pm 2 ab = (a \pm b)^{2}$ . When the top holds $a + b$ (or $a^{2} + b^{2}$ ) and the bottom holds twice the middle term, $2 ab$ (or $2 ab$ ), adding and subtracting produces exact perfect squares. That is what componendo-dividendo does: add on top, subtract below, and the squares appear.

∑ Formula

\frac{a + b}{2 ab} = \frac{p}{q} ⟹ \frac{( a + b ) ^{2}}{( a - b ) ^{2}} = \frac{p + q}{p - q}

The first family

$a, b$: two positive numbers, $a > b$
$p, q$: the numbers of the given ratio

∑ Formula

\frac{a ^{2} + b ^{2}}{2 ab} = \frac{p}{q} ⟹ \frac{( a + b ) ^{2}}{( a - b ) ^{2}} = \frac{p + q}{p - q}

The second family

$a, b$: two positive numbers, $a > b$
$p, q$: the numbers of the given ratio

∑ Formula

x = \frac{2 p m}{m ^{2} + 1} (m > 1) ⟹ \frac{p + x}{p - x} = (\frac{m + 1}{m - 1})^{2}, \frac{p + x - p - x}{p + x + p - x} = \frac{1}{m}

The third family

$p, m$: constants, $m > 1$ so that $p - x > 0$
$x$: the given value

All three share one thought: (1) find where twice the middle term is; (2) make perfect squares by componendo-dividendo; (3) take the root; (4) one more componendo-dividendo gives the ratio you want. In the third family you start by forming $\frac{p}{x} = \frac{m ^{2} + 1}{2 m}$ (dividing by $x$ puts the $2 m$ at the bottom).

Spot itSpot it:

a + b

beside

ab

, or

a^{2} + b^{2}

beside

ab

, or

x

shaped like

\frac{2 \cdot \cdot}{^{2} + 1}

. The test: is twice the middle term present?

Steps

Write the given ratio so that $2 ab$ or $2 ab$ is in the bottom (halve or double both sides if needed).
Componendo-dividendo: add on top, subtract below; both become perfect squares.
Take the square root (the positive root, since $a > b$ ).
One more componendo-dividendo (the way back) gives $\frac{a}{b}$ or the asked fraction.

Example 22 · Basic

(a + b) : ab = 5 : 2

and

a > b > 0

, find

a : b

Solution, step by step

1Bring $2 ab$ into the bottom: $\frac{a + b}{ab} = \frac{5}{2}$ ⟹ $\frac{a + b}{2 ab} = \frac{5}{4}$ .
2Componendo-dividendo: $\frac{( a + b ) ^{2}}{( a - b ) ^{2}} = \frac{5 + 4}{5 - 4} = 9$ ; the square root gives $\frac{a + b}{a - b} = 3$ .
3The way back: $\frac{a}{b} = \frac{3 + 1}{3 - 1} = 2$ , so $\frac{a}{b} = 4$ , that is $a : b = 4 : 1$ .
4Check: $a = 4, b = 1$ gives $(a + b) : ab = 5 : 2$ ✓. Another route: $\frac{a}{b} + \frac{b}{a} = \frac{( a + b ) ^{2}}{ab} - 2 = \frac{25}{4} - 2 = \frac{17}{4}$ , which is $4 + \frac{1}{4}$ .

Answer $4 : 1$

Do not write

a : b = 2 : 1

: it is

a : b = 2 : 1

, and

a : b

is its square.

Example 23 · Standard

\frac{a ^{2} + b ^{2}}{2 ab} = \frac{13}{12}

and

a > b > 0

, find

\frac{a}{b}

Solution, step by step

1The bottom already has $2 ab$ . Componendo-dividendo: $\frac{( a + b ) ^{2}}{( a - b ) ^{2}} = \frac{13 + 12}{13 - 12} = 25$ .
2Square root: $\frac{a + b}{a - b} = 5$ (since $a > b$ both are positive).
3The way back: $\frac{a}{b} = \frac{5 + 1}{5 - 1} = \frac{3}{2}$ . Check: at $a = 3, b = 2$ , $\frac{9 + 4}{12} = \frac{13}{12}$ ✓.

Answer $\frac{3}{2}$

Without the condition

a > b

\frac{a}{b} = \frac{2}{3}

would be equally correct (

a

and

b

swapped).

Example 24 · Standard

x = \frac{2 \times 5 \times 3}{3 ^{2} + 1}

, find the value of

\frac{5 + x - 5 - x}{5 + x + 5 - x}

Solution, step by step

1 $x = \frac{30}{10} = 3$ , and it has the form $\frac{2 p m}{m ^{2} + 1}$ with $p = 5$ , $m = 3$ .
2 $\frac{p}{x} = \frac{m ^{2} + 1}{2 m}$ ; componendo-dividendo: $\frac{p + x}{p - x} = \frac{( m + 1 ) ^{2}}{( m - 1 ) ^{2}} = \frac{16}{4} = 4$ . Square root: $\frac{p + x}{p - x} = 2$ .
3One more componendo-dividendo: $\frac{p + x - p - x}{p + x + p - x} = \frac{2 - 1}{2 + 1} = \frac{1}{3}$ .
4Direct check: $8 - 2 = 22 - 2 = 2$ and $8 + 2 = 32$ , giving $\frac{1}{3}$ ✓. (The answer is always $\frac{1}{m}$ .)

Answer $\frac{1}{3}$

m > 1

is needed; for

m < 1

the same

x

arises (

m

and

\frac{1}{m}

give one

x

) but the answer becomes

m

Example 25 · Advanced

(a + b) : ab = 13 : 6

and

a > b > 0

, find

a : b

Solution, step by step

1With $2 ab$ in the bottom: $\frac{a + b}{2 ab} = \frac{13}{12}$ .
2Componendo-dividendo: $\frac{( a + b ) ^{2}}{( a - b ) ^{2}} = \frac{13 + 12}{13 - 12} = 25$ , so $\frac{a + b}{a - b} = 5$ .
3The way back: $\frac{a}{b} = \frac{5 + 1}{5 - 1} = \frac{3}{2}$ , then square: $\frac{a}{b} = \frac{9}{4}$ , that is $a : b = 9 : 4$ .

Answer $9 : 4$

Check: at

a = 9, b = 4

(a + b) : ab = 13 : 6

✓.

BasicTry it yourself first

(a + b) : ab = 10 : 3

and

a > b > 0

, find

a : b

StandardTry it yourself first

(a + b) : ab = 17 : 4

and

a > b > 0

, find

a : b

StandardTry it yourself first

(a + b) : ab = 29 : 10

and

a > b > 0

, find

a : b

AdvancedTry it yourself first

(a + b) : ab = 34 : 15

and

a > b > 0

, find

a : b

AdvancedTry it yourself first

(a + b) : ab = 53 : 14

and

a > b > 0

, find

a : b

BasicTry it yourself first

\frac{a ^{2} + b ^{2}}{2 ab} = \frac{17}{8}

and

a > b > 0

, find

\frac{a}{b}

StandardTry it yourself first

\frac{a ^{2} + b ^{2}}{2 ab} = \frac{29}{20}

and

a > b > 0

, find

\frac{a}{b}

StandardTry it yourself first

\frac{a ^{2} + b ^{2}}{2 ab} = \frac{13}{5}

and

a > b > 0

, find

\frac{a}{b}

AdvancedTry it yourself first

\frac{a ^{2} + b ^{2}}{2 ab} = \frac{65}{56}

and

a > b > 0

, find

\frac{a}{b}

StandardTry it yourself first

x = \frac{48}{5}

, find the value of

\frac{12 + x}{12 - x}

StandardTry it yourself first

x = \frac{24}{5}

, find the value of

\frac{8 + x - 8 - x}{8 + x + 8 - x}

AdvancedTry it yourself first

x = \frac{80}{17}

, find the value of

\frac{10 + x + 10 - x}{10 + x - 10 - x}

7. When it does not apply, and common slips

Conditions · zero bottoms · the right pair

The conditions for componendo-dividendo: two equal ratios, no zero denominators, the same move on both sides.

The idea

Componendo-dividendo is a rule about an equality, not about a lone fraction. A single $\frac{a}{b}$ gives nothing new because no other fraction is equal to it. But if $\frac{a}{b} = 5$ is given, that is $\frac{a}{b} = \frac{5}{1}$ : an equality exists and the rule applies.

Situation	Allowed?	Reason
$\frac{a}{b} = \frac{c}{d}$ is given	Yes	an equality exists
$\frac{a}{b} = 7$ is given	Yes	$7 = \frac{7}{1}$
A block is an expression, like $\frac{x + y}{z} = \frac{5}{2}$	Yes	take $A = x + y$ , $B = z$
Only $\frac{a}{b}$ with no equality	No	no equal fraction to compare
$\frac{x}{y} = 1$ (that is $x = y$ )	Not here	$x - y = 0$ : zero denominator
A sum of fractions $\frac{a}{b} + \frac{c}{d}$	No	it is an addition, not an equality

Here $A, B$ = the top and bottom blocks and $a, b, c, d$ = the four terms of an equality.

Four common slips: (1) not doing the same move on both sides (on the left $a + b$ but on the right $c - d$ on top). (2) writing the same letters on the right instead of the numbers of its top and bottom. (3) using it when $a - b = 0$ (zero denominator). (4) when the numbers are small the $k$ method is more direct; do not force the rule everywhere.

CheckAfter every use, run a small check: take a pair like

A = 5 k

B = 3 k

and evaluate both sides. If they are not equal, there is a slip somewhere.

Steps

Ask: are there two equal ratios (or one ratio and a number)?
Ask: is any subtraction $A - B$ zero?
Do the same move on both sides and check.

Example 26 · Basic

\frac{x}{y} = 1

, what can you say about

\frac{x + y}{x - y}

Solution, step by step

1By the rule $\frac{x + y}{x - y} = \frac{1 + 1}{1 - 1} = \frac{2}{0}$ : a zero denominator.
2In fact $x = y$ , so $x - y = 0$ and $\frac{x + y}{x - y}$ is not defined at all.
3Lesson: the rule needs $A \neq = B$ ; here work directly with $x = y$ .

Answer undefined (zero denominator)

Example 27 · Standard

\frac{x + y}{z} = \frac{5}{2}

, find the value of

\frac{x + y - z}{x + y + z}

Solution, step by step

1Blocks may be expressions: $A = x + y$ , $B = z$ . Then $\frac{A}{B} = \frac{5}{2}$ is given and $\frac{A - B}{A + B}$ is asked.
2Componendo-dividendo: $\frac{A - B}{A + B} = \frac{5 - 2}{5 + 2} = \frac{3}{7}$ .
3Check: with $x + y = 5, z = 2$ , $\frac{5 - 2}{5 + 2} = \frac{3}{7}$ ✓.

Answer $\frac{3}{7}$

Put brackets in your mind:

(x + y)

is a single unit.

Example 28 · Standard

\frac{a}{b} = \frac{c}{d}

(

b \neq = d

), is

\frac{a + c}{b + d} = \frac{a - c}{b - d}

always true?

Solution, step by step

1Take $\frac{a}{b} = \frac{c}{d} = k$ , so $a = bk$ and $c = d k$ .
2 $\frac{a + c}{b + d} = \frac{k ( b + d )}{b + d} = k$ and $\frac{a - c}{b - d} = \frac{k ( b - d )}{b - d} = k$ ; both equal $k$ .
3So yes, it is true; the only conditions are $b + d \neq = 0$ and $b - d \neq = 0$ .

Answer Yes, it is true

This is a property of equal ratios, not of componendo-dividendo; the next group is about it.

BasicTry it yourself first

\frac{x}{y} = 1

, what is

\frac{x + y}{x - y}

BasicTry it yourself first

Which statement follows from

\frac{a}{b} = \frac{c}{d}

StandardTry it yourself first

From

\frac{p}{q} = \frac{3}{5}

a student wrote

\frac{p + q}{p - q} = \frac{3 + q}{5 - q}

. What is the mistake?

StandardTry it yourself first

In which of these situations does componendo-dividendo not apply directly?

AdvancedTry it yourself first

\frac{a}{b} = \frac{c}{d}

, which is always correct? (all denominators non-zero)

AdvancedTry it yourself first

For which ratio

\frac{a}{b}

\frac{a + b}{a - b}

not defined?

Equal ratios

Put

k

and it all follows

Several equal ratios, and expressions that depend only on a ratio.

8. Equal ratios and the $k$ method

\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k

Giving equal ratios one common value $k$ , and getting the same ratio from the sums of the tops and bottoms.

The idea

When $\frac{a}{b} = \frac{c}{d} = \frac{e}{f}$ , all three fractions have one value; call it $k$ . Then $a = bk$ , $c = d k$ , $e = f k$ . In any expression put $bk, d k, f k$ for $a, c, e$ and $k$ comes out as a factor. This gives the most useful property: the sum of all tops over the sum of all bottoms is also $k$ , because $a + c + e = k (b + d + f)$ .

◭ Diagram

Every row (3 : 2, 6 : 4, 9 : 6) has one ratio. Adding all of them gives

18 : 12 = 3 : 2

again. Adding does not change the ratio.

∑ Formula

\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = k ⟹ \frac{p a + q c + re}{p b + q d + r f} = k, \frac{a ^{2} + c ^{2} + e ^{2}}{b ^{2} + d ^{2} + f ^{2}} = k^{2}

Properties from the

k

method

$k$: the common value of the three fractions
$p, q, r$: any multipliers (as long as the bottom is not zero, e.g. $p b + q d + r f \neq = 0$ )

A chain: if $\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k$ then $b = c k$ , $a = c k^{2}$ and $\frac{a}{d} = k^{3}$ (in $\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{d}$ the middle terms cancel). Sums of pairs: if $\frac{a + b}{p} = \frac{b + c}{q} = \frac{c + a}{r}$ , adding the three gives $a + b + c$ , then subtract one pair at a time.

A hidden trap: in $\frac{x}{y + z} = \frac{y}{z + x} = \frac{z}{x + y} = k$ , adding everything gives $k = \frac{x + y + z}{2 ( x + y + z )} = \frac{1}{2}$ — but only when $x + y + z \neq = 0$ . If $x + y + z = 0$ then $y + z = - x$ and $k = - 1$ . So the answer is $\frac{1}{2}$ or $- 1$ , both.

Spot itSpot it: three or more equal fractions, or a pair such as “

\frac{x}{3} = \frac{y}{5}

”. The first step is always “

= k

”.

Steps

Set every equal fraction equal to $k$ .
Write each variable in terms of $k$ ( $x = 3 k$ and so on).
Put into the given relation and find $k$ .
Build the asked value from $k$ ; if only a ratio is asked, $k$ cancels.

Example 29 · Basic

\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{3}{7}

and

b + d + f = 70

, find

a + c + e

Solution, step by step

1All three fractions equal $\frac{3}{7}$ , so $a = \frac{3}{7} b$ , $c = \frac{3}{7} d$ , $e = \frac{3}{7} f$ .
2Adding, $a + c + e = \frac{3}{7} (b + d + f) = \frac{3}{7} \times 70 = 30$ .

Answer $30$

Three separate tops were unknown, but only their total was needed, so nothing had to be solved.

Example 30 · Standard

\frac{x}{2} = \frac{y}{3} = \frac{z}{7}

and

3 x + 2 y - z = 15

, find

x + y + z

Solution, step by step

1Common value $k$ : $x = 2 k$ , $y = 3 k$ , $z = 7 k$ .
2Put into the condition: $3 (2 k) + 2 (3 k) - 7 k = 6 k + 6 k - 7 k = 5 k = 15$ , so $k = 3$ .
3Now $x = 6$ , $y = 9$ , $z = 21$ and $x + y + z = 36$ .

Answer $36$

Example 31 · Standard

\frac{a + b}{5} = \frac{b + c}{6} = \frac{c + a}{7}

, find

a : b : c

Solution, step by step

1 $a + b = 5 k$ , $b + c = 6 k$ , $c + a = 7 k$ . Adding all three, $2 (a + b + c) = 18 k$ , so $a + b + c = 9 k$ .
2Subtract each pair from the total: $c = 9 k - 5 k = 4 k$ , $a = 9 k - 6 k = 3 k$ , $b = 9 k - 7 k = 2 k$ .
3So $a : b : c = 3 : 2 : 4$ . Check: $a + b = 5 k$ ✓, $b + c = 6 k$ ✓, $c + a = 7 k$ ✓.

Answer $3 : 2 : 4$

Example 32 · Advanced

\frac{x}{y + z} = \frac{y}{z + x} = \frac{z}{x + y} = k

, find all possible values of

k

Solution, step by step

1Adding all tops and all bottoms: $k = \frac{x + y + z}{2 ( x + y + z )}$ .
2Case 1: if $x + y + z \neq = 0$ we may cancel it, so $k = \frac{1}{2}$ .
3Case 2: if $x + y + z = 0$ then $y + z = - x$ , so $\frac{x}{y + z} = \frac{x}{- x} = - 1$ and $k = - 1$ .
4Check: at $x = y = z = 1$ , $k = \frac{1}{2}$ ; at $x = 2, y = 3, z = - 5$ every fraction is $- 1$ ✓.

Answer $k = \frac{1}{2}$ or $k = - 1$

Cancelling

x + y + z

without thinking is exactly the trap here.

BasicTry it yourself first

\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{5}{9}

and

b + d + f = 63

, find

a + c + e

BasicTry it yourself first

\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{4}{11}

and

b + d + f = 66

, find

a + c + e

StandardTry it yourself first

\frac{x}{3} = \frac{y}{4} = \frac{z}{5}

and

2 x + y + z = 60

, find

x + y + z

StandardTry it yourself first

\frac{x}{2} = \frac{y}{3} = \frac{z}{6}

and

3 x - 2 y + z = 18

, find

x + y + z

StandardTry it yourself first

\frac{a + b}{7} = \frac{b + c}{8} = \frac{c + a}{9}

, find

a : b : c

StandardTry it yourself first

\frac{a + b}{9} = \frac{b + c}{10} = \frac{c + a}{11}

, find

a : b : c

AdvancedTry it yourself first

\frac{a + b}{11} = \frac{b + c}{12} = \frac{c + a}{13}

, find

a : b : c

StandardTry it yourself first

\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{2}{5}

, find

a : d

AdvancedTry it yourself first

\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = \frac{3}{4}

, find

a : d

AdvancedTry it yourself first

\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{3}{4}

, find the value of

\frac{a ^{2} + c ^{2} + e ^{2}}{b ^{2} + d ^{2} + f ^{2}}

AdvancedTry it yourself first

\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b}

and

a + b + c \neq = 0

, find the value of each fraction.

9. Homogeneous expressions: put the ratio numbers

Only the ratio matters

When every term has the same total degree, the value depends only on the ratio: put the ratio numbers straight in.

The idea

In $\frac{a ^{2} + c ^{2}}{b ^{2}}$ every term has total degree 2. If you multiply $a, b, c$ all by one number $λ$ , the top and the bottom both become $λ^{2}$ times larger and the fraction does not change. Such an expression is called homogeneous. So once $a : b : c = 6 : 9 : 4$ is known you may put $a = 6, b = 9, c = 4$ directly; no $k$ is needed.

Expression	Homogeneous?	Why
$\frac{a ^{2} + b c}{ab + c ^{2}}$	Yes (degree 2)	every term on top and bottom has degree 2
$\frac{a + b}{a - b}$	Yes (degree 1)	every term has degree 1
$\frac{a + 1}{b + 1}$	No	the number 1 has degree 0, $a$ has 1
$a + b$	Yes, but not a ratio	it changes with $λ$ ; the ratio cannot give it

Here: $a, b, c$ are the variables and $λ$ is any non-zero multiplier. Note that being homogeneous is not enough: a fraction is fixed by the ratio only when the top and the bottom have the same degree. $a + b$ alone is homogeneous but has degree 1 with nothing below, so it depends on $λ$ .

Joining two ratios: given $a : b = 3 : 4$ and $b : c = 6 : 5$ make the shared term $b$ equal. The LCM of 4 and 6 is 12: $a : b = 9 : 12$ and $b : c = 12 : 10$ , so $a : b : c = 9 : 12 : 10$ .

Spot itSpot it: a ratio is given (or can be found) and the asked expression has equal degree on top and bottom. Where a number like

+ 1

appears, this method stops; you need the real values.

Steps

Build $a : b : c$ from the given ratio(s) (equalise the shared term).
Check the degree: is every term on top and bottom of equal degree?
If yes, put the ratio numbers in place of $a, b, c$ and calculate.

Example 33 · Basic

a : b = 2 : 3

and

b : c = 9 : 4

, find the value of

\frac{a ^{2} + b ^{2} + c ^{2}}{ab + b c + c a}

Solution, step by step

1Equalise $b$ : the LCM of 3 and 9 is 9. $a : b = 2 : 3 = 6 : 9$ and $b : c = 9 : 4$ , so $a : b : c = 6 : 9 : 4$ .
2Every term on top and bottom has degree 2, so we may put $a = 6, b = 9, c = 4$ .
3Top $36 + 81 + 16 = 133$ ; bottom $54 + 36 + 24 = 114$ .
4 $\frac{133}{114} = \frac{7 \times 19}{6 \times 19} = \frac{7}{6}$ .

Answer $\frac{7}{6}$

The ratio numbers may differ from the real values, but here the answer stays the same.

Example 34 · Standard

\frac{3 x + 2 y}{3 x - 2 y} = \frac{11}{3}

, find the value of

\frac{x ^{2} + y ^{2}}{x y}

Solution, step by step

1Componendo-dividendo (blocks $3 x$ and $2 y$ ): $\frac{3 x}{2 y} = \frac{11 + 3}{11 - 3} = \frac{7}{4}$ , so $\frac{x}{y} = \frac{7}{4} \times \frac{2}{3} = \frac{7}{6}$ .
2The asked expression is homogeneous (every term has degree 2), so put $x = 7, y = 6$ .
3 $\frac{49 + 36}{42} = \frac{85}{42}$ .

Answer $\frac{85}{42}$

Example 35 · Standard

x : y = 2 : 3

, can the value of

\frac{x + 2}{y + 2}

be found? And what if also

x + y = 20

Solution, step by step

1 $\frac{x + 2}{y + 2}$ is not homogeneous (the number 2 has degree 0). With $x = 2 k, y = 3 k$ it becomes $\frac{2 k + 2}{3 k + 2}$ , which depends on $k$ ; the ratio alone does not fix the value.
2Now $x + y = 20$ is also given: $5 k = 20$ , $k = 4$ , so $x = 8, y = 12$ .
3Then $\frac{x + 2}{y + 2} = \frac{10}{14} = \frac{5}{7}$ .

Answer not from the ratio alone; with $x + y = 20$ it is $\frac{5}{7}$

The moment a plain number is added, you need the real values.

Example 36 · Advanced

2 a = 3 b = 4 c

, find the value of

\frac{a + b}{b + c}

Solution, step by step

1Set all three equal to $12 k$ (the LCM of 2, 3, 4 is 12): $a = 6 k$ , $b = 4 k$ , $c = 3 k$ , that is $a : b : c = 6 : 4 : 3$ .
2The expression is homogeneous (degree 1 on top and 1 below): $\frac{6 + 4}{4 + 3} = \frac{10}{7}$ .

Answer $\frac{10}{7}$

A bigger multiplier does not mean a bigger variable: the one with the larger coefficient is the smaller number.

BasicTry it yourself first

a : b = 3 : 4

and

b : c = 6 : 5

, find the value of

\frac{a + c}{b}

BasicTry it yourself first

a : b = 2 : 5

and

b : c = 10 : 3

, find the value of

\frac{a ^{2} + b ^{2} + c ^{2}}{ab + b c + c a}

StandardTry it yourself first

a : b = 4 : 3

and

b : c = 9 : 7

, find the value of

\frac{ab + b c}{b ^{2} + c a}

StandardTry it yourself first

a : b = 5 : 2

and

b : c = 6 : 7

, find the value of

\frac{a ^{2} - c ^{2}}{b ^{2}}

AdvancedTry it yourself first

a : b = 7 : 6

and

b : c = 4 : 9

, find the value of

\frac{2 a - b + 3 c}{a + b - c}

AdvancedTry it yourself first

a : b = 2 : 3

and

b : c = 6 : 5

, find the value of

\frac{a}{b} + \frac{b}{c} + \frac{c}{a}

StandardTry it yourself first

3 a = 4 b = 6 c

, find the value of

\frac{a + c}{b}

AdvancedTry it yourself first

5 a = 6 b = 10 c

, find the value of

\frac{a ^{2} + b ^{2} + c ^{2}}{ab + b c + c a}

AdvancedTry it yourself first

Which of these can be found from the ratio

x : y = 3 : 5

alone?

A condition that shrinks the expression

+1,

a + b + c = 0

ab c = 1

When a condition is given ( $a + b + c = 0$ , $pq + q r + r p = 0$ , $ab c = 1$ ) or fractions look like $k$ and $x + k$ , small moves open the whole question.

10. The add-1 / subtract-1 move

\frac{x}{x + k} = 1 - \frac{k}{x + k}

When one sum of fractions is given and a sum with the same bottoms is asked, add or subtract 1 in every term.

The idea

The engine of componendo was “add 1 to both sides”. The same move on a sum of fractions shrinks long questions. The core fact: $\frac{x}{x + k} + \frac{k}{x + k} = \frac{x + k}{x + k} = 1$ . A whole 1 is split into two parts: one part is given in the question and the other is asked. Knowing one, the other is “1 minus the rest”.

◭ Diagram

Four bars, each of length 1. If the saffron parts (

\frac{k}{x + k}

) add up to

S

, the indigo parts (

\frac{x}{x + k}

) add up to

n - S

, here

4 - S

∑ Formula

\sum \frac{k}{x + k} = S ⟹ \sum \frac{x}{x + k} = n - S, \sum \frac{k}{x - k} = S ⟹ \sum \frac{x}{x - k} = n + S

The two rules, with

+ 1

and with

- 1

$n$: the number of terms
$S$: the given sum
$k$: the own constant of each term

∑ Formula

\frac{P - Q}{P + Q} + 1 = \frac{2 P}{P + Q}, 1 - \frac{P - Q}{P + Q} = \frac{2 Q}{P + Q}, \frac{a}{b + c} + 1 = \frac{a + b + c}{b + c}

Three more shapes

$P, Q$: the two blocks of the fraction
$a, b, c$: the variables

Spot itSpot it: the given sum and the asked sum have the same bottoms but different tops (

k

x

P - Q

P

Steps

Compare the bottoms of the given and the asked fractions: are they the same?
See whether the asked term is made from the given one by $+ 1$ or by $1 -$ .
Count the terms, $n$ .
Add $n$ to the total or take it from $n$ ; divide by 2 if needed.

Example 37 · Basic

\frac{2}{x + 2} + \frac{4}{x + 4} + \frac{6}{x + 6} + \dots + \frac{40}{x + 40} = 25

, find the value of

\frac{x}{x + 2} + \frac{x}{x + 4} + \dots + \frac{x}{x + 40}

Solution, step by step

1The terms use $2, 4, 6, \dots, 40$ , so there are $\frac{40}{2} = 20$ terms.
2For each term $\frac{x}{x + k} = 1 - \frac{k}{x + k}$ (because $\frac{x}{x + k} + \frac{k}{x + k} = 1$ ).
3For the 20 terms: $20 - 25 = - 5$ .

Answer $- 5$

A negative answer need not alarm you;

x

is such that some terms exceed 1.

Example 38 · Standard

\frac{3}{x - 3} + \frac{5}{y - 5} + \frac{7}{z - 7} = 11

, find the value of

\frac{x}{x - 3} + \frac{y}{y - 5} + \frac{z}{z - 7}

Solution, step by step

1There is a minus in the bottom, so $\frac{x}{x - 3} = \frac{( x - 3 ) + 3}{x - 3} = 1 + \frac{3}{x - 3}$ : here 1 is added.
2The same in all three terms: total $= 3 + 11 = 14$ .

Answer $14$

Add or subtract depends on the sign in the bottom: subtract for

x + k

, add for

x - k

Example 39 · Standard

\frac{m ^{2} - 3 n}{m ^{2} + 3 n} + \frac{n ^{2} - 3 k}{n ^{2} + 3 k} + \frac{k ^{2} - 3 m}{k ^{2} + 3 m} = \frac{1}{2}

, find the value of

\frac{m ^{2}}{m ^{2} + 3 n} + \frac{n ^{2}}{n ^{2} + 3 k} + \frac{k ^{2}}{k ^{2} + 3 m}

Solution, step by step

1Each term has the form $\frac{P - Q}{P + Q}$ ( $P = m^{2}$ , $Q = 3 n$ and so on). Adding 1: $\frac{P - Q}{P + Q} + 1 = \frac{2 P}{P + Q}$ .
2Adding 1 to each of the three terms, $2 \times (the asked sum) = \frac{1}{2} + 3 = \frac{7}{2}$ .
3So the asked sum is $\frac{7}{4}$ .

Answer $\frac{7}{4}$

Do not forget to halve: the term is

\frac{2 P}{P + Q}

, not

\frac{P}{P + Q}

Example 40 · Advanced

\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = 5

, find the value of

(a + b + c) (\frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b})

Solution, step by step

1Add 1 to the first term: $\frac{a}{b + c} + 1 = \frac{a + ( b + c )}{b + c} = \frac{a + b + c}{b + c}$ .
2The same for the second and third terms; adding 1 to all three the left side becomes $\frac{a + b + c}{b + c} + \frac{a + b + c}{c + a} + \frac{a + b + c}{a + b}$ .
3The right side is $5 + 3 = 8$ . Taking $(a + b + c)$ out, the asked expression is $8$ .

Answer $8$

The reverse also works: if

(a + b + c) \sum \frac{1}{b + c} = 8

then

\sum \frac{a}{b + c} = 5

BasicTry it yourself first

\frac{3}{x + 3} + \frac{6}{x + 6} + \dots + \frac{36}{x + 36} = 20

, find the value of

\frac{x}{x + 3} + \frac{x}{x + 6} + \dots + \frac{x}{x + 36}

StandardTry it yourself first

\frac{5}{x + 5} + \frac{10}{x + 10} + \dots + \frac{40}{x + 40} = \frac{27}{2}

, find the value of

\frac{x}{x + 5} + \frac{x}{x + 10} + \dots + \frac{x}{x + 40}

BasicTry it yourself first

\frac{2}{x - 2} + \frac{6}{y - 6} = 9

, find the value of

\frac{x}{x - 2} + \frac{y}{y - 6}

StandardTry it yourself first

\frac{4}{x - 4} + \frac{9}{y - 9} + \frac{11}{z - 11} + \frac{13}{w - 13} = 16

, find the value of

\frac{x}{x - 4} + \frac{y}{y - 9} + \frac{z}{z - 11} + \frac{w}{w - 13}

StandardTry it yourself first

\frac{p - 2 q}{p + 2 q} + \frac{q - 2 r}{q + 2 r} + \frac{r - 2 p}{r + 2 p} = \frac{1}{3}

, find the value of

\frac{p}{p + 2 q} + \frac{q}{q + 2 r} + \frac{r}{r + 2 p}

StandardTry it yourself first

\frac{p - 3 q}{p + 3 q} + \frac{q - 3 r}{q + 3 r} + \frac{r - 3 p}{r + 3 p} = \frac{3}{2}

, find the value of

\frac{3 q}{p + 3 q} + \frac{3 r}{q + 3 r} + \frac{3 p}{r + 3 p}

StandardTry it yourself first

\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = 7

, find the value of

(a + b + c) (\frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b})

AdvancedTry it yourself first

\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = 11

, find the value of

(a + b + c) (\frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b})

StandardTry it yourself first

\frac{a}{a + b} + \frac{b}{b + c} + \frac{c}{c + a} = \frac{5}{2}

, find the value of

\frac{b}{a + b} + \frac{c}{b + c} + \frac{a}{c + a}

AdvancedTry it yourself first

\frac{a}{a + b} + \frac{b}{b + c} + \frac{c}{c + a} = 4

, find the value of

\frac{b}{a + b} + \frac{c}{b + c} + \frac{a}{c + a}

11. When $a + b + c = 0$

a^{3} + b^{3} + c^{3} = 3 ab c

When three quantities add up to zero, each pair sums to the negative of the third and the sum of cubes becomes $3 ab c$ .

The idea

When three quantities add up to zero, each is the negative of the sum of the other two: $a + b = - c$ , $b + c = - a$ , $c + a = - b$ . Everything else follows. For instance $c^{2} = (a + b)^{2}$ , so $a^{2} + b^{2} - c^{2} = - 2 ab$ . And in the cube identity $a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$ the first bracket is zero, so $a^{3} + b^{3} + c^{3} = 3 ab c$ .

∑ Formula

a + b + c = 0 ⟹ ⎩ ⎨ ⎧ a^{3} + b^{3} + c^{3} = 3 ab c a^{2} + b^{2} + c^{2} = - 2 (ab + b c + c a) (a + b) (b + c) (c + a) = - ab c

Results of

a + b + c = 0

$a, b, c$: three numbers or expressions whose sum is zero

A hidden zero-sum is the biggest treasure. Wherever you see three quantities, add them: (1) $(a - b) + (b - c) + (c - a) = 0$ always, so $(a - b)^{3} + (b - c)^{3} + (c - a)^{3} = 3 (a - b) (b - c) (c - a)$ . (2) In numbers, such as $23 + (- 9) + (- 14) = 0$ . (3) If $a + b = c$ then $a + b + (- c) = 0$ , so $a^{3} + b^{3} - c^{3} = - 3 ab c$ .

Spot itSpot it: “

a + b + c = 0

” is given, or three cubes/squares whose bases add to 0, or three cyclic differences.

Steps

Add the three quantities: is the sum zero?
If yes, pick the right result from above.
If it is not visible, replace each pair by the negative of the third and bring the expression down to one letter.
Check the answer with $a = 1, b = 2, c = - 3$ (this set satisfies $a + b + c = 0$ ).

Example 41 · Basic

Find the value of

2 8^{3} - 1 3^{3} - 1 5^{3}

Solution, step by step

1Add the bases with their signs: $28 + (- 13) + (- 15) = 0$ .
2The sum is zero, so the sum of cubes is $3 \times 28 \times (- 13) \times (- 15)$ .
3 $13 \times 15 = 195$ , $195 \times 28 = 5460$ , $\times 3 = 16380$ .
4The long way, as a check: $21952 - 2197 - 3375 = 16380$ ✓.

Answer $16380$

A cube keeps the sign of its base:

- 1 3^{3} = (- 13)^{3}

Example 42 · Standard

a + b + c = 0

, find the value of

\frac{( a ^{2} + b ^{2} + c ^{2} ) ^{2}}{a ^{4} + b ^{4} + c ^{4}}

Solution, step by step

1From $a + b + c = 0$ , $ab + b c + c a = - \frac{1}{2} (a^{2} + b^{2} + c^{2})$ . Write $s = a^{2} + b^{2} + c^{2}$ .
2 $a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = (ab + b c + c a)^{2} - 2 ab c (a + b + c) = \frac{s ^{2}}{4} - 0$ .
3 $a^{4} + b^{4} + c^{4} = s^{2} - 2 \cdot \frac{s ^{2}}{4} = \frac{s ^{2}}{2}$ .
4So the fraction is $\frac{s ^{2}}{s ^{2} /2} = 2$ . Check: at $(1, 2, - 3)$ , $\frac{1 4 ^{2}}{1 + 16 + 81} = \frac{196}{98} = 2$ ✓.

Answer $2$

Example 43 · Standard

For

a = 17, b = 12, c = 9

, find

(a - b)^{3} + (b - c)^{3} + (c - a)^{3}

Solution, step by step

1The three brackets: $a - b = 5$ , $b - c = 3$ , $c - a = - 8$ .
2Their sum is $5 + 3 - 8 = 0$ , so the sum of cubes $= 3 \times 5 \times 3 \times (- 8) = - 360$ .
3Check: $125 + 27 - 512 = - 360$ ✓.

Answer $- 360$

Cyclic differences always sum to zero, whatever

a, b, c

are.

Example 44 · Advanced

a + b + c = 0

and

a^{2} + b^{2} + c^{2} = 14

, find

a^{4} + b^{4} + c^{4}

Solution, step by step

1 $ab + b c + c a = - \frac{14}{2} = - 7$ (because $0 = (a + b + c)^{2} = 14 + 2 (ab + b c + c a)$ ).
2 $a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = (ab + b c + c a)^{2} - 2 ab c (a + b + c) = 49$ .
3 $a^{4} + b^{4} + c^{4} = 1 4^{2} - 2 \times 49 = 196 - 98 = 98$ .
4Check: at $(1, 2, - 3)$ , $1 + 16 + 81 = 98$ ✓.

Answer $98$

Straight from

\frac{s ^{2}}{2}

\frac{196}{2} = 98

BasicTry it yourself first

Find the value of

4 7^{3} - 2 1^{3} - 2 6^{3}

StandardTry it yourself first

Find the value of

3 5^{3} - 1 8^{3} - 1 7^{3}

StandardTry it yourself first

For

p = 25, q = 14, r = 9

, find

(p - q)^{3} + (q - r)^{3} + (r - p)^{3}

StandardTry it yourself first

For

p = 40, q = 23, r = 31

, find

(p - q)^{3} + (q - r)^{3} + (r - p)^{3}

StandardTry it yourself first

Find the value of

5^{3} + 6^{3} - 1 1^{3}

AdvancedTry it yourself first

Find the value of

8^{3} + 9^{3} - 1 7^{3}

AdvancedTry it yourself first

a + b + c = 0

and

a^{2} + b^{2} + c^{2} = 26

, find

a^{4} + b^{4} + c^{4}

BasicTry it yourself first

a + b + c = 0

, find the value of

\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}

StandardTry it yourself first

a + b + c = 0

, find the value of

\frac{a ^{2}}{b c} + \frac{b ^{2}}{c a} + \frac{c ^{2}}{ab}

StandardTry it yourself first

a + b + c = 0

, find the value of

\frac{( a + b ) ( b + c ) ( c + a )}{ab c}

AdvancedTry it yourself first

a + b + c = 0

, find the value of

\frac{a ^{2} + b ^{2} + c ^{2}}{ab + b c + c a}

AdvancedTry it yourself first

a + b + c = 0

, find the value of

\frac{a ^{2} + b ^{2} - c ^{2}}{ab}

12. $pq + q r + r p = 0$ and $ab c = 1$

Turn the bottoms into a shared one

Both conditions turn the bottoms of the fractions into the same expression or multiples of one another.

The idea

First condition: if $x y + yz + z x = 0$ then $yz = - x (y + z)$ . So $x^{2} - yz = x^{2} + x y + x z = x (x + y + z)$ : the awkward bottom gets the common factor $s = x + y + z$ ; likewise $y^{2} - z x = y s$ and $z^{2} - x y = z s$ . After that everything simplifies.

Second condition: if $ab c = 1$ , the three terms $\frac{1}{1 + a + ab}$ , $\frac{1}{1 + b + b c}$ , $\frac{1}{1 + c + c a}$ can be brought to one bottom. Multiply numerator and denominator of the second by $a$ : $\frac{a}{a + ab + ab c} = \frac{a}{a + ab + 1}$ . Multiply numerator and denominator of the third by $ab$ : $\frac{ab}{ab + ab c + a ^{2} b c} = \frac{ab}{ab + 1 + a}$ . All three bottoms are now $1 + a + ab$ and the tops add to $1 + a + ab$ : the sum is $1$ .

∑ Formula

x y + yz + z x = 0 ⟹ x^{2} - yz = x (x + y + z), ab c = 1 ⟹ \frac{1}{1 + a + ab} + \frac{1}{1 + b + b c} + \frac{1}{1 + c + c a} = 1

The results of the two conditions

$x, y, z, a, b, c$: non-zero numbers
$s$: $x + y + z$ (not zero)

Remember: $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$ is also the same as $ab + b c + c a = 0$ (common bottom $ab c$ ), so this method applies there as well.

Spot itSpot it: bottoms like

x^{2} - yz

(a square minus the product of the other two) or cyclic bottoms such as

1 + a + ab

, and the condition

x y + yz + z x = 0

ab c = 1

Steps

Write the condition to suit the bottoms (for instance $- yz = x y + x z$ ).
Take out the common factor $s$ from each bottom; if $ab c = 1$ multiply term by term as needed.
The terms are now simple: add them.
Check with a set that satisfies the condition (for $x y + yz + z x = 0$ take $x = y = 2, z = - 1$ ).

Example 45 · Standard

x y + yz + z x = 0

, find the value of

\frac{yz}{x ^{2} - yz} + \frac{z x}{y ^{2} - z x} + \frac{x y}{z ^{2} - x y}

Solution, step by step

1 $- yz = x y + x z$ , so $x^{2} - yz = x (x + y + z) = x s$ ; in the same way $y^{2} - z x = ys$ and $z^{2} - x y = zs$ .
2The sum $= \frac{yz}{x s} + \frac{z x}{ys} + \frac{x y}{zs} = \frac{1}{s} \cdot \frac{( yz ) ^{2} + ( z x ) ^{2} + ( x y ) ^{2}}{x yz}$ (common bottom $x yz$ ).
3 $0 = (x y + yz + z x)^{2} = (x y)^{2} + (yz)^{2} + (z x)^{2} + 2 x yz (x + y + z)$ , so $(x y)^{2} + (yz)^{2} + (z x)^{2} = - 2 x yz s$ .
4The sum $= \frac{1}{s} \cdot \frac{- 2 x yz s}{x yz} = - 2$ . Check: at $x = y = 2, z = - 1$ , $\frac{- 2}{6} + \frac{- 2}{6} + \frac{4}{- 3} = - 2$ ✓.

Answer $- 2$

s = 0

the bottoms would be zero, so the question assumes

s \neq = 0

Example 46 · Standard

x y + yz + z x = 0

, find the value of

\frac{x ^{2} + yz}{x ^{2} - yz} + \frac{y ^{2} + z x}{y ^{2} - z x} + \frac{z ^{2} + x y}{z ^{2} - x y}

Solution, step by step

1Bottom: $x^{2} - yz = x s$ (as before). Top: $x^{2} + yz = x^{2} - x y - x z = x (x - y - z)$ .
2The first term is $\frac{x ( x - y - z )}{x s} = \frac{x - y - z}{s} = \frac{2 x - s}{s}$ (because $x - y - z = 2 x - s$ ).
3Add the three terms: $\frac{2 ( x + y + z ) - 3 s}{s} = \frac{2 s - 3 s}{s} = - 1$ .

Answer $- 1$

Example 47 · Standard

ab c = 1

, find the value of

\frac{1}{1 + a + ab} + \frac{1}{1 + b + b c} + \frac{1}{1 + c + c a}

Solution, step by step

1Keep the first term as it is: $\frac{1}{1 + a + ab}$ .
2Multiply numerator and denominator of the second term by $a$ : $\frac{a}{a + ab + ab c} = \frac{a}{a + ab + 1}$ (because $ab c = 1$ ).
3Multiply numerator and denominator of the third term by $ab$ : $\frac{ab}{ab + ab c + a ^{2} b c} = \frac{ab}{ab + 1 + a}$ ( $ab c = 1$ and $a^{2} b c = a$ ).
4All three bottoms are $1 + a + ab$ , so the sum is $\frac{1 + a + ab}{1 + a + ab} = 1$ .
5Check: at $a = 2, b = 3, c = \frac{1}{6}$ , $\frac{1}{9} + \frac{2}{9} + \frac{2}{3} = 1$ ✓.

Answer $1$

a = b = c = 1

also gives 1, but that is a weak check; the second set (

2, 3, \frac{1}{6}

) is more trustworthy.

Example 48 · Advanced

\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0

, find the value of

\frac{b c}{a ^{2}} + \frac{c a}{b ^{2}} + \frac{ab}{c ^{2}}

Solution, step by step

1Let $u = \frac{1}{a}$ , $v = \frac{1}{b}$ , $w = \frac{1}{c}$ : then $u + v + w = 0$ .
2 $\frac{b c}{a ^{2}} = \frac{u ^{2}}{v w}$ (since $\frac{1}{a ^{2}} = u^{2}$ and $b c = \frac{1}{v w}$ ). So the sum $= \frac{u ^{2}}{v w} + \frac{v ^{2}}{w u} + \frac{w ^{2}}{uv} = \frac{u ^{3} + v ^{3} + w ^{3}}{uv w}$ .
3From $u + v + w = 0$ , $u^{3} + v^{3} + w^{3} = 3 uv w$ , so the sum is $3$ .

Answer $3$

Inverting to make a new variable (here

u = \frac{1}{a}

) is a good move.

BasicTry it yourself first

x y + yz + z x = 0

, find the value of

\frac{x}{x ^{2} - yz} + \frac{y}{y ^{2} - z x} + \frac{z}{z ^{2} - x y}

StandardTry it yourself first

x y + yz + z x = 0

, find the value of

\frac{yz}{x ^{2} - yz} + \frac{z x}{y ^{2} - z x} + \frac{x y}{z ^{2} - x y}

AdvancedTry it yourself first

x y + yz + z x = 0

, find the value of

\frac{x ^{2} + yz}{x ^{2} - yz} + \frac{y ^{2} + z x}{y ^{2} - z x} + \frac{z ^{2} + x y}{z ^{2} - x y}

StandardTry it yourself first

ab c = 1

, find the value of

\frac{a}{1 + a + ab} + \frac{b}{1 + b + b c} + \frac{c}{1 + c + c a}

StandardTry it yourself first

For

a = 2, b = 3, c = \frac{1}{6}

, find

\frac{1}{1 + a + ab} + \frac{1}{1 + b + b c} + \frac{1}{1 + c + c a}

by direct counting.

13. Cyclic conditions: $x + \frac{1}{y} = 1$

Back in three steps

From $x + \frac{1}{y} = 1$ and $y + \frac{1}{z} = 1$ it follows that $z + \frac{1}{x} = 1$ and $x yz = - 1$ .

The idea

$x + \frac{1}{y} = 1$ means $y = \frac{1}{1 - x}$ : once $x$ is known, $y$ is known. The same rule applied from $y$ to $z$ gives $z = \frac{1}{1 - y}$ . Tying it back to $x$ we get $z = \frac{x - 1}{x}$ , and then $z + \frac{1}{x} = \frac{x - 1}{x} + \frac{1}{x} = 1$ . So in three steps the same rule goes the full circle and comes back to the start.

◭ Diagram

Every arrow is the same rule (

next = \frac{1}{1 - previous}

). After three arrows

x

becomes

x

again.

∑ Formula

x + \frac{1}{y} = 1, y + \frac{1}{z} = 1 ⟹ z + \frac{1}{x} = 1, x yz = - 1

The result of the cycle

$x, y, z$: three non-zero numbers, $x, y \neq = 1$

Why $x yz = - 1$ ? $y = \frac{1}{1 - x}$ and $z = \frac{x - 1}{x}$ . Multiplying, $x yz = x \cdot \frac{1}{1 - x} \cdot \frac{x - 1}{x} = \frac{x - 1}{1 - x} = - 1$ . A special caution: this miracle happens only when the sum is $1$ . Take a condition like $x + \frac{1}{y} = 2$ and the value of $z + \frac{1}{x}$ changes with $y$ (example 4 below); there even “putting values” will not give a single answer.

Spot itSpot it: three variables, each equation with one variable and the reciprocal of the next, going round; the sum (or value) is

1

Steps

From one equation write the next variable: $y = \frac{1}{1 - x}$ .
Repeat the same rule: $z = \frac{1}{1 - y}$ .
Check the third equation or the product.

Example 49 · Basic

x + \frac{1}{y} = 1

y + \frac{1}{z} = 1

and

x = 3

, find

y

and

z

and check

z + \frac{1}{x}

Solution, step by step

1First condition: $\frac{1}{y} = 1 - 3 = - 2$ , so $y = - \frac{1}{2}$ .
2Second: $\frac{1}{z} = 1 - y = 1 + \frac{1}{2} = \frac{3}{2}$ , so $z = \frac{2}{3}$ .
3Check: $z + \frac{1}{x} = \frac{2}{3} + \frac{1}{3} = 1$ ✓ and $x yz = 3 \times (- \frac{1}{2}) \times \frac{2}{3} = - 1$ ✓.

Answer $y = - \frac{1}{2}, z = \frac{2}{3}$

None of the three is 0 or 1; otherwise some fraction would have a zero bottom.

Example 50 · Standard

x + \frac{1}{y} = 1

and

y + \frac{1}{z} = 1

, find the value of

x yz

Solution, step by step

1From the first condition $x = 1 - \frac{1}{y} = \frac{y - 1}{y}$ .
2From the second condition $\frac{1}{z} = 1 - y$ , so $z = \frac{1}{1 - y}$ .
3Multiplying, $x yz = \frac{y - 1}{y} \cdot y \cdot \frac{1}{1 - y} = \frac{y - 1}{1 - y} = - 1$ .

Answer $- 1$

\frac{y - 1}{1 - y}

the top and bottom are negatives of each other, hence

- 1

Example 51 · Standard

x + \frac{1}{y} = 1

and

y + \frac{1}{z} = 1

, find the value of

x y + yz + z x - (x + y + z)

Solution, step by step

1Multiplying the first condition by $y$ gives $x y + 1 = y$ , that is $x y = y - 1$ .
2Multiplying the second condition by $z$ gives $yz = z - 1$ . Multiplying the third ( $z + \frac{1}{x} = 1$ ) by $x$ gives $z x = x - 1$ .
3Add the three: $x y + yz + z x = (y - 1) + (z - 1) + (x - 1) = x + y + z - 3$ .
4Therefore $x y + yz + z x - (x + y + z) = - 3$ .

Answer $- 3$

The third condition came from the result of the cycle; without it this step would not work.

Example 52 · Advanced

x + \frac{1}{y} = 2

and

y + \frac{1}{z} = 2

, does

z + \frac{1}{x}

have one fixed value?

Solution, step by step

1Take $y = 3$ : $x = 2 - \frac{1}{3} = \frac{5}{3}$ and $\frac{1}{z} = 2 - 3 = - 1$ , so $z = - 1$ . Then $z + \frac{1}{x} = - 1 + \frac{3}{5} = - \frac{2}{5}$ .
2Take $y = 4$ : $x = 2 - \frac{1}{4} = \frac{7}{4}$ and $\frac{1}{z} = 2 - 4 = - 2$ , so $z = - \frac{1}{2}$ . Then $z + \frac{1}{x} = - \frac{1}{2} + \frac{4}{7} = \frac{1}{14}$ .
3Two valid sets gave two different answers ( $- \frac{2}{5}$ and $\frac{1}{14}$ ), so the value is not fixed.

Answer No; the value changes with $y$

This is the basic rule of value putting: if two valid sets give different answers, the question has no single answer.

BasicTry it yourself first

x + \frac{1}{y} = 1

y + \frac{1}{z} = 1

and

x = 5

, find

z

BasicTry it yourself first

x + \frac{1}{y} = 1

y + \frac{1}{z} = 1

and

x = - 2

, find

z

StandardTry it yourself first

x + \frac{1}{y} = 1

y + \frac{1}{z} = 1

and

y = 4

, find

x + z

StandardTry it yourself first

x + \frac{1}{y} = 1

y + \frac{1}{z} = 1

and

y = - 3

, find

x + z

AdvancedTry it yourself first

x + \frac{1}{y} = 1

y + \frac{1}{z} = 1

and

x = \frac{2}{3}

, find

z

StandardTry it yourself first

x + \frac{1}{y} = 1

and

y + \frac{1}{z} = 1

, what is

z + \frac{1}{x}

AdvancedTry it yourself first

x + \frac{1}{y} = 1

and

y + \frac{1}{z} = 1

, what is

x^{2} y^{2} z^{2}

Value putting

Put numbers, cross out options

Putting convenient numbers for letters, testing options, respecting conditions and spotting where the method cheats; plus remainder-factor shortcuts, symmetric expressions and minimum-maximum.

14. The idea, when it is valid, choosing numbers

One legal set is enough

An identity is true for every value, so one convenient legal set tells you the answer or the wrong options.

The idea

An identity is an equality that holds for every value of the letters, like $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ . So if the answer to a question is a fixed identity, put $a = 1, b = 2$ and count both sides: they must agree. An option that fails this check is certainly wrong. This is value putting: replace letters by numbers and turn algebra into arithmetic.

◭ Diagram

Four moves. If more than one option survives the fourth, go round again with other numbers.

There are three kinds of questions. (1) An identity: no condition (“which of these is equal?”): any legal numbers. (2) A condition: such as $a + b + c = 0$ : the numbers must satisfy it (next method). (3) Find $x$ : put numbers for the letters ( $a, b$ ), get one numerical value of $x$ and test the options with it.

Six rules for choosing numbers: (1) small whole numbers, so the arithmetic is done in the head. (2) Different letters get different numbers ( $a = 1, b = 2, c = 3$ ), because equal numbers hide differences. (3) Not $0$ or $1$ as a first choice ( $0$ wipes out terms and $1$ makes all powers equal). (4) No zero denominator in any fraction; look at the denominators of the options too. (5) Satisfy the conditions of the question ( $a > b$ , $x > 0$ , $a + b + c = 0$ ). (6) Choose by looking at the options: if they differ in $a^{2}$ against $a^{4}$ , do not put $a = 1$ .

Condition	A good first set	Check / remark
none	$a = 1, b = 2, c = 3$	all different and small
$a + b + c = 0$	$1, 2, - 3$	sum $0$ ✓ (beware $1, 1, - 2$ : several denominators become zero)
$ab + b c + c a = 0$	$2, 2, - 1$	$4 - 2 - 2 = 0$ ✓
$ab c = 1$	$2, 3, \frac{1}{6}$	product $1$ ✓ ( $1, 1, 1$ is a weak test)
$\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$	$3, 6, 2$	$\frac{1}{3} + \frac{1}{6} = \frac{1}{2}$ ✓
$a^{2} + b^{2} = c^{2}$	$3, 4, 5$	$9 + 16 = 25$ ✓
$\frac{a}{b} = \frac{c}{d}$	$6, 3, 4, 2$	$\frac{6}{3} = \frac{4}{2}$ ✓
$a + b = k$	any $a$ , then $b = k - a$	choose one letter, derive the other

Here: $a, b, c, d$ are the letters of the question and $k$ is a given constant. The sets in the table are only first tries; choose another set for the second test.

Spot itSpot it: four options and an expression in letters; or “find the value” where the answer must be a number (

a + b + c = 0

and so on). Caution: value putting proves nothing; it knocks out wrong options. When it cheats is covered in “When value putting fails”.

Steps

Identify: an identity, a condition question, or finding $x$ .
Choose numbers by the six rules; if there is a condition, check it first.
Evaluate the asked expression.
Evaluate every option at the same numbers; cross out the ones that do not match.
If more than one is left, test only the survivors with new numbers.

Example 53 · Basic

(a + b + c)^{3} - a^{3} - b^{3} - c^{3}

equals which of these? (A)

3 (a + b) (b + c) (c + a)

(B)

24 ab c

(C)

(a + b) (b + c) (c + a)

(D)

3 (a - b) (b - c) (c - a)

Solution, step by step

1This is an identity (no condition). Choose small different numbers: $a = 1, b = 2, c = 3$ .
2The asked expression: $(1 + 2 + 3)^{3} - 1 - 8 - 27 = 216 - 36 = 180$ .
3(A) $3 \times (3) (5) (4) = 180$ ✓. (B) $24 \times 6 = 144$ ✗. (C) $3 \times 5 \times 4 = 60$ ✗. (D) $3 \times (- 1) (- 1) (2) = 6$ ✗.
4Had we put $a = b = c = 1$ the asked value would be $27 - 3 = 24$ and both (A) and (B) would give $24$ : the choice would be left open. Different numbers prevented that trap.

Answer (A) $3 (a + b) (b + c) (c + a)$

Example 54 · Standard

a + b + c = 0

, find the value of

\frac{a ^{2}}{a ^{2} - b c} + \frac{b ^{2}}{b ^{2} - c a} + \frac{c ^{2}}{c ^{2} - ab}

Solution, step by step

1The condition is $a + b + c = 0$ , so choose a set that sums to zero: $a = 1, b = 2, c = - 3$ ( $1 + 2 - 3 = 0$ ✓).
2Check the bottoms: $a^{2} - b c = 1 + 6 = 7$ , $b^{2} - c a = 4 + 3 = 7$ , $c^{2} - ab = 9 - 2 = 7$ ; none is zero ✓.
3The sum $= \frac{1}{7} + \frac{4}{7} + \frac{9}{7} = \frac{14}{7} = 2$ .
4To be sure, use $(2, 3, - 5)$ : every bottom is $4 + 15 = 19$ ; the tops add to $4 + 9 + 25 = 38$ ; $\frac{38}{19} = 2$ ✓.

Answer $2$

By algebra: when

a + b + c = 0

the three bottoms are equal (

\frac{a ^{2} + b ^{2} + c ^{2}}{2}

), and the fraction is

\frac{a ^{2} + b ^{2} + c ^{2}}{( a ^{2} + b ^{2} + c ^{2} ) /2} = 2

Example 55 · Standard

What is

x

\frac{x - a}{b} + \frac{x - b}{a} = \frac{a}{b} + \frac{b}{a}

? (A)

a + b

(B)

\frac{a ^{2} + b ^{2}}{a + b}

(C)

\frac{2 ( a ^{2} + b ^{2} )}{a + b}

(D)

ab

Solution, step by step

1Here $x$ is to be found. Put numbers for the letters: $a = 2, b = 3$ (the denominators $a, b$ are not zero ✓).
2The equation $\frac{x - 2}{3} + \frac{x - 3}{2} = \frac{2}{3} + \frac{3}{2}$ : multiply by 6: $2 (x - 2) + 3 (x - 3) = 4 + 9$ , that is $5 x - 13 = 13$ , so $x = \frac{26}{5}$ .
3Options: (A) $5$ ✗, (B) $\frac{13}{5}$ ✗, (C) $\frac{2 \times 13}{5} = \frac{26}{5}$ ✓, (D) $6$ ✗.

Answer (C) $\frac{2 ( a ^{2} + b ^{2} )}{a + b}$

By algebra:

a (x - a) + b (x - b) = a^{2} + b^{2}

⟹

(a + b) x = 2 (a^{2} + b^{2})

Example 56 · Advanced

x = a (b - c)

y = b (c - a)

z = c (a - b)

, then

(\frac{x}{a})^{3} + (\frac{y}{b})^{3} + (\frac{z}{c})^{3}

equals: (A)

\frac{3 x yz}{ab c}

(B)

\frac{x yz}{ab c}

(C)

x + y + z

(D)

0

Solution, step by step

1 $a = 1, b = 2, c = 4$ (different; $a, b, c \neq = 0$ ✓). Then $x = 1 (2 - 4) = - 2$ , $y = 2 (4 - 1) = 6$ , $z = 4 (1 - 2) = - 4$ .
2 $\frac{x}{a} = - 2$ , $\frac{y}{b} = 3$ , $\frac{z}{c} = - 1$ ; the sum of cubes is $- 8 + 27 - 1 = 18$ .
3(A) $\frac{3 \times ( - 2 ) \times 6 \times ( - 4 )}{8} = \frac{144}{8} = 18$ ✓. (B) $6$ ✗. (C) $- 2 + 6 - 4 = 0$ ✗. (D) $0$ ✗.
4Why: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = (b - c) + (c - a) + (a - b) = 0$ , three quantities with sum zero, so the sum of cubes is three times their product.

Answer (A) $\frac{3 x yz}{ab c}$

0

, but

18 \neq = 0

, so both are out.

BasicTry it yourself first

Which of these is equal to

(a + b)^{3} - (a - b)^{3}

BasicTry it yourself first

Which of these is equal to

\frac{x ^{4} + x ^{2} + 1}{x ^{2} + x + 1}

StandardTry it yourself first

Which of these is equal to

(a + b + c) (ab + b c + c a) - ab c

StandardTry it yourself first

Which of these is equal to

\frac{a ^{3} + b ^{3} + c ^{3} - 3 ab c}{a + b + c}

StandardTry it yourself first

Which of these is equal to

(x + y)^{3} - x^{3} - y^{3}

AdvancedTry it yourself first

Which of these is equal to

(a^{2} + b^{2}) (c^{2} + d^{2}) - (a c + b d)^{2}

AdvancedTry it yourself first

Which of these is equal to

(a + b + c)^{2} + (a + b - c)^{2} + (a - b + c)^{2} + (b + c - a)^{2}

15. Testing options: numbers that separate them

Target first, then options · the two-test rule

In an options question get the asked value once and match every option against it; if two options survive, take new numbers.

The idea

In an options question the aim is not to construct the right answer but to cross out the wrong options. The biggest mistake is to choose numbers that make several options look right at once. Choose the numbers by looking at the options so that every option gives a different value. Then the order: (1) get the asked value once; (2) get every option at the same numbers; (3) whatever does not match is out; (4) if more than one is left, new numbers, tested on the survivors only.

◭ Diagram

The trial table of example 2: the first set keeps all four, the second cuts three.

How the options differ	What to avoid	Why
options differ in power ( $a^{2}$ , $a^{4}$ )	avoid $a = 0$ and $a = 1$	every power gives just $0$ or $1$
options differ in sign ( $a + b$ , $- (a + b)$ )	keep $a + b \neq = 0$	the negative of zero is zero
symmetric forms ( $a + b + c$ , $ab + b c + c a$ , $a^{2} + b^{2} + c^{2}$ )	avoid $a = b = c = 1$	all three give $3$
options with factors like $(a - b)$	keep $a \neq = b$	otherwise the factor becomes $0$
options with $\frac{1}{a}$ , $\frac{1}{b}$	$a, b \neq = 0$	no zero denominator

Here: $a, b, c$ are the letters of the question. This table is rule (6) of the previous method, spelled out for each kind of option.

The two-test rule: when exactly one option wins, that is often enough; but if time allows, test the same option once more with other numbers. If “none of these” is among the options, refute each of A, B, C with at least one legal set before choosing it; one option looking right once does not make it right.

HabitWrite the value of every option at every set (as in the table): an option that fails at one set is out for good.

Steps

Read the options and see what separates them (power, sign, factor).
Use the table to avoid the value that would make them equal; take other small different numbers.
Get the asked value and the value of every option.
Test the survivor on other numbers.

Example 57 · Standard

For

(x + 2 b) (x + 4 b) (x + 6 b) (x + 8 b) + k

to be a perfect square, what is

k

? (A)

4 b^{4}

(B)

16 b^{2}

(C)

16 b^{4}

(D)

8 b^{4}

Solution, step by step

1This is an identity: $k$ must make a perfect square for every $x, b$ . The options differ in $b^{2}$ against $b^{4}$ , so do not put $b = 1$ (then $b^{2} = b^{4}$ and (B), (C) would both be $16$ ).
2For convenience take $x = 0$ , $b = 2$ : the terms are $4, 8, 12, 16$ and the product is $4 c d o t 8 c d o t 12 c d o t 16 = 32 im es 192 = 6144$ .
3At $b = 2$ the options are: (A) $4 im es 16 = 64$ , (B) $16 im es 4 = 64$ , (C) $16 im es 16 = 256$ , (D) $8 im es 16 = 128$ .
4 $6144 + 64 = 6208$ is not a square ( $7 8^{2} = 6084$ , $7 9^{2} = 6241$ ) ✗; $6144 + 256 = 6400 = 8 0^{2}$ ✓; $6144 + 128 = 6272$ is not a square ( $7 9^{2} = 6241$ , $8 0^{2} = 6400$ ) ✗.
5Only (C) survives: $k = 16 b^{4}$ . By algebra: pair them, $(x + 2 b) (x + 8 b) = x^{2} + 10 b x + 16 b^{2}$ and $(x + 4 b) (x + 6 b) = x^{2} + 10 b x + 24 b^{2}$ ; with $y = x^{2} + 10 b x + 20 b^{2}$ the product is $(y - 4 b^{2}) (y + 4 b^{2}) = y^{2} - 16 b^{4}$ , so $k = 16 b^{4}$ ✓.

Answer (C) $16 b^{4}$

The numerical values of (A) and (B) happened to coincide (64); both were wrong anyway. Two options having equal values is not a worry as long as both differ from the right value.

Example 58 · Standard

\frac{x - a}{b + c} + \frac{x - b}{c + a} + \frac{x - c}{a + b} = 3

, then

x =

? (A)

a + b + c

(B)

ab + b c + c a

(C)

a^{2} + b^{2} + c^{2}

(D)

\frac{( a + b + c ) ^{2}}{3}

Solution, step by step

1The easiest set is $a = b = c = 1$ : bottoms $b + c = 2$ and so on; the equation is $3 \cdot \frac{x - 1}{2} = 3$ , so $x = 3$ .
2At $a = b = c = 1$ all four options give $3$ : (A) $3$ , (B) $3$ , (C) $3$ , (D) $\frac{9}{3} = 3$ . Nothing is cut; this set is useless (the price of choosing all equal).
3Now different numbers: $a = 1, b = 2, c = 3$ , bottoms $5, 4, 3$ . The equation $\frac{x - 1}{5} + \frac{x - 2}{4} + \frac{x - 3}{3} = 3$ ; multiply by 60: $12 (x - 1) + 15 (x - 2) + 20 (x - 3) = 180$ , that is $47 x - 102 = 180$ , $x = 6$ .
4Options: (A) $6$ ✓; (B) $2 + 3 + 6 = 11$ ✗; (C) $1 + 4 + 9 = 14$ ✗; (D) $\frac{36}{3} = 12$ ✗.

Answer (A) $a + b + c$

The trial table is in the figure above. By algebra: with

x = a + b + c

each term is

\frac{b + c}{b + c} = 1

Example 59 · Basic

Which is the factorisation of

x^{3} + 6 x^{2} + 11 x + 6

? (A)

(x + 1) (x + 2) (x + 4)

(B)

(x + 1) (x + 2) (x + 3)

(C)

(x + 1) (x + 3) (x + 4)

(D)

(x - 1) (x + 2) (x + 3)

Solution, step by step

1The brackets differ, so choose $x = 1$ (it makes the $x - 1$ option $0$ and gives the others different products).
2The polynomial: $1 + 6 + 11 + 6 = 24$ .
3(A) $2 \cdot 3 \cdot 5 = 30$ ✗; (B) $2 \cdot 3 \cdot 4 = 24$ ✓; (C) $2 \cdot 4 \cdot 5 = 40$ ✗; (D) $0$ ✗.
4Second check at $x = 2$ : the polynomial is $8 + 24 + 22 + 6 = 60$ and (B) is $3 \cdot 4 \cdot 5 = 60$ ✓.

Answer (B) $(x + 1) (x + 2) (x + 3)$

Sometimes one test settles it; do a second one if time allows.

BasicTry it yourself first

Which of these is equal to

(a - b)^{2} + (b - c)^{2} + (c - a)^{2}

BasicTry it yourself first

Which of these is equal to

(a^{2} - b^{2})^{2} + (2 ab)^{2}

StandardTry it yourself first

Which of these is equal to

(a + b)^{2} (a - b)^{2}

StandardTry it yourself first

Which of these is equal to

\frac{1}{ab} + \frac{1}{b c} + \frac{1}{c a}

StandardTry it yourself first

Which of these is equal to

\frac{a ^{2} - b ^{2}}{( a - b ) ^{2}} + \frac{a + b}{a - b}

AdvancedTry it yourself first

Which of these is equal to

(a + b + c)^{3} - (a + b - c)^{3} - (b + c - a)^{3} - (c + a - b)^{3}

AdvancedTry it yourself first

Which of these is equal to

(a + b) (b + c) (c + a) + ab c

16. Putting numbers under a condition

Choose the set, check the condition

With a condition ( $a + b + c = 0$ , $ab c = 1$ , …) build numbers that satisfy it: choose all but one and get the last from the condition.

The idea

In a condition question not every number will do, only a set that meets the condition. That is also its strength: the condition has made the answer fixed, so any legal set gives the same answer. The trick to build a set: choose all letters but one and get the last from the condition (usually a linear equation).

Condition	Choose → solve	Set
$a + b + c = 0$	$a = 1, b = 2$ ⟹ $c = - 3$	$(1, 2, - 3)$
$ab + b c + c a = 0$	$a = 2, b = 2$ ⟹ $4 + 2 c + 2 c = 0$ ⟹ $c = - 1$	$(2, 2, - 1)$
$ab c = 1$	$a = 2, b = 3$ ⟹ $c = \frac{1}{6}$	$(2, 3, \frac{1}{6})$
$\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$	$a = 3, b = 6$ ⟹ $\frac{1}{c} = \frac{1}{2}$	$(3, 6, 2)$
$a = b + c$	$b = 2, c = 3$ ⟹ $a = 5$	$(5, 2, 3)$

Here: the set in the right column meets the condition on the left; you can build another set of your own. The checking rule: after building a set, put the numbers into the condition and make sure it holds; first the set, then the check, then the question.

Spot itSpot it: “if ..., find the value of ...” where the answer is a number or a simple expression. Caution: with two conditions both must hold; and choose letters so that no denominator becomes zero.

Steps

Read the condition; choose values for all letters but one (small and different).
Get the last letter from the condition.
Check: does the condition hold? Is any denominator zero?
Get the asked value. If time allows, confirm with a second set.

Example 60 · Standard

a + b + c = 0

, find the value of

\frac{a ^{2}}{2 a ^{2} + b c} + \frac{b ^{2}}{2 b ^{2} + c a} + \frac{c ^{2}}{2 c ^{2} + ab}

Solution, step by step

1The easiest set is $(1, 1, - 2)$ : the sum is $0$ ✓. But look at the bottoms: $2 a^{2} + b c = 2 + (1) (- 2) = 0$ , a zero denominator, so this set is forbidden.
2Another set $(1, 2, - 3)$ : the sum is $0$ ✓. Bottoms: $2 + (2) (- 3) = - 4$ ; $2 \cdot 4 + (- 3) (1) = 5$ ; $2 \cdot 9 + (1) (2) = 20$ , all non-zero ✓.
3The sum $= \frac{1}{- 4} + \frac{4}{5} + \frac{9}{20} = \frac{- 5 + 16 + 9}{20} = \frac{20}{20} = 1$ .
4Another set $(2, 3, - 5)$ : bottoms $8 - 15 = - 7$ ; $18 - 10 = 8$ ; $50 + 6 = 56$ ; the sum $= \frac{4}{- 7} + \frac{9}{8} + \frac{25}{56} = \frac{- 32 + 63 + 25}{56} = 1$ ✓.

Answer $1$

Lesson: the “easiest set” can turn out to be forbidden. The bottom

2 a^{2} + b c = (a - b) (a - c)

here, so no two of

a, b, c

may be equal.

Example 61 · Standard

x + y = 10

and

x y = 21

, find

x^{3} + y^{3}

Solution, step by step

1There are two conditions, so the set must satisfy both. $x + y = 10$ and $x y = 21$ mean that $x, y$ are the two numbers with sum $10$ and product $21$ : $3$ and $7$ .
2Check: $3 + 7 = 10$ ✓ and $3 \times 7 = 21$ ✓.
3 $x^{3} + y^{3} = 27 + 343 = 370$ .
4By algebra: $(x + y)^{3} - 3 x y (x + y) = 1000 - 630 = 370$ ✓.

Answer $370$

The formula and value putting give the same answer; but with value putting you must find the set that fulfils both conditions.

Example 62 · Standard

ab + b c + c a = 0

, find the value of

\frac{( a + b ) ( b + c ) ( c + a )}{ab c}

Solution, step by step

1For the condition choose $a = 2, b = 2$ and solve for $c$ : $4 + 2 c + 2 c = 0$ ⟹ $c = - 1$ . Check: $4 - 2 - 2 = 0$ ✓.
2Top: $(2 + 2) (2 - 1) (- 1 + 2) = 4 \cdot 1 \cdot 1 = 4$ ; bottom: $2 \cdot 2 \cdot (- 1) = - 4$ .
3The fraction is $\frac{4}{- 4} = - 1$ .
4To be sure take $(3, 6, - 2)$ : $18 - 12 - 6 = 0$ ✓; top $9 \cdot 4 \cdot 1 = 36$ , bottom $3 \cdot 6 \cdot (- 2) = - 36$ ; $- 1$ ✓.

Answer $- 1$

By algebra:

(a + b) (b + c) (c + a) = (a + b + c) (ab + b c + c a) - ab c = - ab c

Example 63 · Advanced

a + b + c = 6

and

ab + b c + c a = 11

, find the value of

a^{3} + b^{3} + c^{3} - 3 ab c

Solution, step by step

1Two conditions: the sum is $6$ and the sum of pair products is $11$ . The numbers must meet both: $1, 2, 3$ ( $1 + 2 + 3 = 6$ ✓ and $2 + 6 + 3 = 11$ ✓).
2 $1 + 8 + 27 - 3 \cdot 6 = 36 - 18 = 18$ .
3By algebra: $(a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a) = 6 \times (36 - 22 - 11) = 6 \times 3 = 18$ ✓.

Answer $18$

A set meeting only the first condition (such as

2, 2, 2

) would give the wrong answer

0

here; the next method is about exactly this.

BasicTry it yourself first

a + b = 1

, find the value of

a^{3} + b^{3} + 3 ab

BasicTry it yourself first

a - b = 2

, find the value of

a^{3} - b^{3} - 6 ab

StandardTry it yourself first

\frac{a}{b} = \frac{c}{d}

, find the value of

\frac{( a ^{2} + c ^{2} ) ( b ^{2} + d ^{2} )}{( ab + c d ) ^{2}}

StandardTry it yourself first

a + b + c = 0

, find the value of

\frac{a ^{2} + b ^{2} + c ^{2}}{a ^{2} - b c}

StandardTry it yourself first

a = b + c

, find the value of

a^{3} - b^{3} - c^{3} - 3 ab c

AdvancedTry it yourself first

a + b + c = 0

, find the value of

\frac{( a - b ) ^{2} + ( b - c ) ^{2} + ( c - a ) ^{2}}{a ^{2} + b ^{2} + c ^{2}}

AdvancedTry it yourself first

ab + b c + c a = 0

, find the value of

\frac{a ^{2} + b ^{2} + c ^{2}}{( a + b + c ) ^{2}}

AdvancedTry it yourself first

a + b + c = 6, ab + b c + c a = 11

, find the value of

a^{3} + b^{3} + c^{3} - 3 ab c

17. When value putting fails

Seven traps and the checking rule

Coincidence, equal values, forbidden values, a half-satisfied condition, non-homogeneous expressions, no fixed answer, and no real set to be found.

The idea

Value putting proves nothing. It can do only two things: (a) show that an option is wrong (it fails at one legal set); (b) offer a candidate for the answer if the answer is a fixed number. When the choice of set is careless, the same method can make a wrong option look right. Below are the seven situations where this happens; exam questions are built around exactly these traps.

Trap	What happens	Cure
1. Coincidence	one set makes two or three options look right	a second set, on the survivors only
2. Equal values	$a = b = c$ or $b = c$ erases terms like $(a - b)$ , $(b - c)$	different values
3. Forbidden values	a zero denominator, or a limit such as $x > 0$ , $a > b$ , $m > 1$ is broken	check bottoms and limits first
4. Half a condition	the set meets only one condition, or the condition was “split equally”	put the set into every condition
5. Not homogeneous	ratio numbers were put in place of real values	find the real values
6. No fixed answer	two legal sets give different answers	the question has no single answer
7. No real set	no simple real solution of the condition exists	use the algebraic identity

Here: a “set” = the numbers chosen for the letters; “legal” = it meets the conditions and limits. The rule for choosing “none of these”: you must show each of A, B, C wrong with some legal set; that all three failed at one set is not enough, since the set itself may be illegal or the arithmetic may slip.

RuleThe checking rule (the two-test rule): (1) check the set against conditions and denominators; (2) evaluate the target and the options; (3) if more than one option survives, use a second set; (4) even a lone survivor deserves one match on a second set.

Steps

The moment you choose a set, check it against every condition, every denominator (the options’ too) and every limit.
Evaluate the target and the options; cross out the mismatches.
If two or more survive, take a new set that separates them.
If two legal sets give different answers, either the question has no fixed answer or an arithmetic slip was made.

Example 64 · Standard

a = \frac{2 b c}{b + c}

, which statement is true? (A)

a = \frac{b + c}{2}

(B)

\frac{2}{a} = \frac{1}{b} + \frac{1}{c}

(C)

a^{2} = b c

(D)

a = b c

Solution, step by step

1First the easiest, equal values: $b = c = 3$ , so $a = \frac{2 \cdot 9}{6} = 3$ .
2(A) $\frac{3 + 3}{2} = 3$ ✓; (B) $\frac{2}{3} = \frac{1}{3} + \frac{1}{3}$ ✓; (C) $9 = 9$ ✓; (D) $3 = 9$ ✓: all four look right. Equal values made the means (AM, GM, HM) coincide.
3Now unequal values $b = 3, c = 6$ : $a = \frac{2 \cdot 18}{9} = 4$ .
4(A) $\frac{9}{2} = 4.5$ ✗; (B) $\frac{2}{4} = \frac{1}{2}$ and $\frac{1}{3} + \frac{1}{6} = \frac{1}{2}$ ✓; (C) $16$ against $18$ ✗; (D) $4$ against $18$ ✗.

Answer (B)

b = c

we get

a = b = c

and every mean is equal, which is why equal values must not be the first choice.

Example 65 · Standard

For

x = \frac{2 p m}{m ^{2} + 1}

(

m > 1

) the claim is

\frac{p + x - p - x}{p + x + p - x} = \frac{1}{m}

. What happens when we test it with

p = 5

and

m = \frac{1}{2}

Solution, step by step

1 $x = \frac{2 \cdot 5 \cdot \frac{1}{2}}{\frac{1}{4} + 1} = \frac{5}{\frac{5}{4}} = 4$ .
2The fraction is $\frac{9 - 1}{9 + 1} = \frac{3 - 1}{3 + 1} = \frac{1}{2}$ , whereas the claim gives $\frac{1}{m} = 2$ : no match.
3But the claim was for $m > 1$ and $m = \frac{1}{2}$ is outside that limit. The same $x = 4$ also comes from $m = 2$ , and then $\frac{1}{m} = \frac{1}{2}$ ✓ matches.

Answer a set outside the limit made a correct claim look wrong

A set that breaks the limit can make value putting cut the right answer, not only the wrong ones.

Example 66 · Standard

a + b + c = 6

and

ab + b c + c a = 11

, find

a^{3} + b^{3} + c^{3} - 3 ab c

. A student put

a = b = c = 2

and got

0

; where is the mistake?

Solution, step by step

1 $a = b = c = 2$ meets the first condition ( $2 + 2 + 2 = 6$ ✓) but not the second: $ab + b c + c a = 12 \neq = 11$ ✗.
2So the set is illegal and the $8 + 8 + 8 - 24 = 0$ it gave is useless.
3A legal set $(1, 2, 3)$ : $6$ ✓ and $11$ ✓; the value is $36 - 18 = 18$ . Algebra: $6 \times (36 - 22 - 11) = 18$ ✓.

Answer $18$

When building the set, check every condition, not just the first.

Example 67 · Advanced

a + b + c = 12

and

\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = \frac{17}{10}

, find

\frac{a ^{2}}{b + c} + \frac{b ^{2}}{c + a} + \frac{c ^{2}}{a + b}

Solution, step by step

1The wrong method (“all three equal”): suppose each fraction is $\frac{17}{30}$ . Then $\frac{a}{12 - a} = \frac{17}{30}$ ⟹ $a = \frac{204}{47}$ and all three numbers are this, but then $a + b + c = \frac{612}{47} \neq = 12$ : the first condition is broken. The answer it gives, $3 \cdot \frac{a ^{2}}{12 - a} = \frac{1734}{235} \approx 7.38$ , cannot be trusted.
2Correct algebra: $\frac{a ^{2}}{b + c} = \frac{a ( a + b + c )}{b + c} - a$ (because $\frac{a ( a + b + c )}{b + c} = \frac{a ^{2}}{b + c} + a$ ). Adding the three: $\sum \frac{a ^{2}}{b + c} = (a + b + c) \sum \frac{a}{b + c} - (a + b + c) = 12 \times \frac{17}{10} - 12$ .
3 $= \frac{102}{5} - 12 = \frac{42}{5}$ .
4Check with a legal set: $(2, 4, 6)$ : the sum is $12$ ✓; $\frac{2}{10} + \frac{4}{8} + \frac{6}{6} = \frac{17}{10}$ ✓. Then $\frac{4}{10} + \frac{16}{8} + \frac{36}{6} = \frac{42}{5}$ ✓.

Answer $\frac{42}{5}$

If the condition were “the sum of the fractions is

1

” the answer would be

(a + b + c) (1 - 1) = 0

and the wrong “all equal” method would also give

0

(its assumption forces

a + b + c = 0

): it looks right by luck now and then. Do not trust it.

Example 68 · Advanced

\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = \frac{1}{2}

, write

\frac{a ^{2}}{b + c} + \frac{b ^{2}}{c + a} + \frac{c ^{2}}{a + b}

in terms of

a + b + c

. Can the result be checked by value putting?

Solution, step by step

1As in the previous example, $\sum \frac{a ^{2}}{b + c} = (a + b + c) (\sum \frac{a}{b + c} - 1) = (a + b + c) (\frac{1}{2} - 1) = - \frac{a + b + c}{2}$ .
2A set is needed for a check. Take $b = c = t$ : $\frac{a}{2 t} + \frac{2 t}{a + t} = \frac{1}{2}$ ⟹ $a^{2} + 3 t^{2} = 0$ , so there is no real $a$ .
3Value putting is then useless; the algebraic formula is the answer (and it holds in every legal case).

Answer $- \frac{a + b + c}{2}$

“No simple real set can be found” is also a trap: then use clever algebra.

BasicTry it yourself first

Which set

(a, b, c)

is legal for

a + b + c = 0

BasicTry it yourself first

The expression has the denominator

(a - b) (b - c)

. Which set

(a, b, c)

is safe?

StandardTry it yourself first

Of four options

A

and

B

both came out right at

a = b = c = 1

. What should you do next?

StandardTry it yourself first

Conditions

a + b + c = 6

and

ab + b c + c a = 11

. Which set meets both?

StandardTry it yourself first

a > b > 0

and the expression contains

a - b

. Which set

(a, b)

is legal?

AdvancedTry it yourself first

The result

\frac{1}{m}

for

x = \frac{2 p m}{m ^{2} + 1}

holds only for

m > 1

. Someone who put

m = \frac{1}{2}

got

\frac{1}{2}

instead of

2

. What was the mistake?

AdvancedTry it yourself first

a + b + c = 15

and

\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = \frac{13}{8}

, find

\frac{a ^{2}}{b + c} + \frac{b ^{2}}{c + a} + \frac{c ^{2}}{a + b}

AdvancedTry it yourself first

x : y = 2 : 3

and

x + y = 10

, what is

\frac{x + 1}{y + 1}

AdvancedTry it yourself first

To separate the statements “

a = \frac{b + c}{2}

” and “

\frac{2}{a} = \frac{1}{b} + \frac{1}{c}

” for

a = \frac{2 b c}{b + c}

, which set

(b, c)

will work?

18. Special values in polynomials: remainder and factor

f (0), f (1), f (- 1), f (a)

Putting $x = 0, 1, - 1$ or the zero of the divisor into a polynomial gives coefficient sums, remainders and factors at once.

The idea

A polynomial $f (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots$ is true for every $x$ like an identity, so chosen values of $x$ make the coefficients speak: $f (0) = a_{0}$ (the constant term); $f (1) = a_{0} + a_{1} + a_{2} + \dots$ (the sum of all coefficients); $f (- 1) = a_{0} - a_{1} + a_{2} - \dots$ (the even-power sum $-$ the odd-power sum). Adding and subtracting the two gives the even-power and odd-power sums separately.

The remainder theorem: when $f (x)$ is divided by $(x - a)$ we write $f (x) = (x - a) q (x) + r$ ; putting $x = a$ leaves $f (a)$ on the left and only $r$ on the right, so $r = f (a)$ . The factor theorem: $(x - a)$ is a factor exactly when $f (a) = 0$ . For a divisor $(p x + q)$ put $x = - \frac{q}{p}$ .

∑ Formula

f (x) = (x - a) q (x) + r ⟹ r = f (a), even sum = \frac{f ( 1 ) + f ( - 1 )}{2}, odd sum = \frac{f ( 1 ) - f ( - 1 )}{2}

Rules for special values

$f (x)$: the given polynomial
$q (x), r$: the quotient and the remainder (the remainder has lower degree than the divisor)
$a$: the zero of the divisor $x - a$

A divisor with two factors, such as $(x - 1) (x - 2)$ : the remainder is $p x + q$ (one degree lower). Putting $x = 1$ and $x = 2$ gives two equations: $p + q = f (1)$ and $2 p + q = f (2)$ ; solve them.

Spot itSpot it: “remainder”, “is it a factor”, “value of

k

so that it divides”, “sum of all coefficients”: skip the long division and put a value.

Steps

Set the divisor equal to zero and find its zero.
Put that value into $f (x)$ : it is the remainder (zero means a factor).
For coefficient-sum questions put $x = 1$ and $x = - 1$ .
If $k$ is unknown, form a linear equation in $k$ from $f (a) = r$ .

Example 69 · Basic

(2 x - 1)^{6} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{6} x^{6}

, find

a_{0} + a_{2} + a_{4} + a_{6}

Solution, step by step

1 $x = 1$ : $a_{0} + a_{1} + \dots + a_{6} = (2 - 1)^{6} = 1$ .
2 $x = - 1$ : $a_{0} - a_{1} + a_{2} - \dots + a_{6} = (- 2 - 1)^{6} = 3^{6} = 729$ .
3Adding the two cancels the odd powers: $2 (a_{0} + a_{2} + a_{4} + a_{6}) = 1 + 729 = 730$ .
4So $a_{0} + a_{2} + a_{4} + a_{6} = 365$ .

Answer $365$

Expanding fully would need seven terms and binomial coefficients.

Example 70 · Standard

When

x^{3} - 4 x^{2} + k x + 6

is divided by

(x - 2)

the remainder is

4

. Find

k

Solution, step by step

1Remainder theorem: the remainder is $f (2)$ .
2 $f (2) = 8 - 16 + 2 k + 6 = 2 k - 2$ .
3 $2 k - 2 = 4$ ⟹ $k = 3$ . Check: $f (2) = 8 - 16 + 6 + 6 = 4$ ✓.

Answer $k = 3$

Example 71 · Standard

Find the remainder when

f (x) = x^{101} - 3 x^{50} + 7

is divided by

x^{2} - 1

Solution, step by step

1The divisor is $(x - 1) (x + 1)$ , so the remainder is $p x + q$ : $f (x) = (x^{2} - 1) g (x) + p x + q$ .
2 $x = 1$ : $f (1) = 1 - 3 + 7 = 5$ , so $p + q = 5$ .
3 $x = - 1$ : $f (- 1) = - 1 - 3 + 7 = 3$ ( $(- 1)^{101} = - 1$ , $(- 1)^{50} = 1$ ), so $- p + q = 3$ .
4Adding, $2 q = 8$ , $q = 4$ ; then $p = 1$ . The remainder is $x + 4$ .

Answer $x + 4$

Example 72 · Standard

For the polynomial

x^{3} + 6 x^{2} + 3 x - 10

test which of

(x - 1)

(x + 2)

and

(x - 3)

are factors.

Solution, step by step

1 $f (1) = 1 + 6 + 3 - 10 = 0$ ✓: $(x - 1)$ is a factor.
2 $f (- 2) = - 8 + 24 - 6 - 10 = 0$ ✓: $(x + 2)$ is a factor. (For $(x + 2)$ we put $x = - 2$ .)
3 $f (3) = 27 + 54 + 9 - 10 = 80 \neq = 0$ ✗: $(x - 3)$ is not a factor; the remainder is $80$ .
4The third factor: from the constant term $- 10 = (- 1) (2) (c)$ we get $c = 5$ , so $(x - 1) (x + 2) (x + 5)$ .

Answer $(x - 1)$ and $(x + 2)$ are factors

Sign care: for

(x + 2)

we put

x = - 2

, not

+ 2

BasicTry it yourself first

Expand

f (x) = (3 x - 2)^{5}

a_{0} + a_{1} x + a_{2} x^{2} + \dots

and find the sum of all the coefficients.

StandardTry it yourself first

Expand

f (x) = (x^{2} + x + 1)^{4}

a_{0} + a_{1} x + a_{2} x^{2} + \dots

and find the sum of all the coefficients.

StandardTry it yourself first

f (x) = (2 x + 1)^{5} = a_{0} + a_{1} x + a_{2} x^{2} + \dots

, find the sum of the coefficients of the even powers.

AdvancedTry it yourself first

f (x) = (x - 3)^{6} = a_{0} + a_{1} x + a_{2} x^{2} + \dots

, find the sum of the coefficients of the odd powers.

BasicTry it yourself first

Find the remainder when

x^{3} - 5 x^{2} + 2 x + 9

is divided by

(x - 3)

StandardTry it yourself first

Find the remainder when

2 x^{4} - x^{3} + 3 x - 4

is divided by

(x + 1)

StandardTry it yourself first

When

x^{3} + k x^{2} - 4 x + 5

is divided by

(x - 2)

the remainder is

9

. Find

k

AdvancedTry it yourself first

When

2 x^{3} - 5 x^{2} + k x - 6

is divided by

(x + 1)

the remainder is

- 15

. Find

k

StandardTry it yourself first

Find the remainder when

6 x^{3} - 7 x^{2} + 2 x + 8

is divided by

(2 x - 1)

AdvancedTry it yourself first

Find the remainder when

x^{5} - 2 x^{2} + 3 x + 1

is divided by

(x - 1) (x - 2)

19. Symmetric and cyclic expressions: factors from zeros

zero at

a = b

⟹

(a - b)

is a factor

An expression that is zero at $a = b$ has the factor $(a - b)$ ; match the degree and put one value to get the constant.

The idea

An expression is symmetric if swapping any two letters leaves it unchanged ( $a^{2} + b^{2} + c^{2}$ , $ab + b c + c a$ ). It is cyclic if replacing $a \to b \to c \to a$ leaves it unchanged ( $a^{2} b + b^{2} c + c^{2} a$ is cyclic but not symmetric). $(a - b) (b - c) (c - a)$ is cyclic and changes sign when two letters swap.

Factors from zeros: the factor theorem works for letters too. If putting $a = b$ makes the expression $0$ , then $(a - b)$ is a factor (just as zero at $x = 2$ made $(x - 2)$ a factor). In a cyclic expression the same holds for $b = c$ and $c = a$ , so $(a - b) (b - c) (c - a)$ is a whole factor. The rest is fixed by degree: if the total degree is 3 only a constant remains; if it is 4 a symmetric factor $k (a + b + c)$ of degree one remains. The constant $k$ is found by putting one value.

∑ Formula

a^{2} (b - c) + b^{2} (c - a) + c^{2} (a - b) a^{3} (b - c) + b^{3} (c - a) + c^{3} (a - b) (a + b + c)^{3} - a^{3} - b^{3} - c^{3} = - (a - b) (b - c) (c - a) = - (a - b) (b - c) (c - a) (a + b + c) = 3 (a + b) (b + c) (c + a)

Three useful factor forms

$a, b, c$: any three numbers or expressions

Spot it: a cyclic expression with differences like $(b - c)$ , $(c - a)$ , $(a - b)$ multiplied in, or a sum of three cubes/squares to be turned into one factor form. Caution: always check these forms on two different sets (in example 1 below one set $(0, 1, 2)$ gives the constant and another checks it).

CheckValue putting works exactly here because both sides are the same identity: to find the constant take a set in which no factor is zero (

0, 1, 2

is good;

1, 1, 2

is bad).

Steps

Put $a = b$ : is the expression zero? Then $(a - b)$ is a factor; if cyclic, so are $(b - c)$ and $(c - a)$ .
Count the total degree and find the degree of the remaining factor.
Find the remaining constant $k$ (or $k (a + b + c)$ ) from one set such as $(0, 1, 2)$ .
Check on a second set.

Example 73 · Standard

Show that

a^{2} (b - c) + b^{2} (c - a) + c^{2} (a - b) = k (a - b) (b - c) (c - a)

and find

k

Solution, step by step

1Put $a = b$ : $b^{2} (b - c) + b^{2} (c - b) + c^{2} \cdot 0 = b^{2} (b - c) - b^{2} (b - c) = 0$ . So $(a - b)$ is a factor.
2The expression is cyclic, so $(b - c)$ and $(c - a)$ are also factors. Their degree is $3$ — the same as the expression — so only a constant $k$ remains.
3For $k$ take $(a, b, c) = (0, 1, 2)$ : the left side $= 0 + 1 \cdot 2 + 4 \cdot (- 1) = - 2$ ; $(a - b) (b - c) (c - a) = (- 1) (- 1) (2) = 2$ . So $k = - 1$ .
4Check with a second set $(1, 2, 4)$ : the left side $= 1 (- 2) + 4 (3) + 16 (- 1) = - 6$ ; the right side $- (- 1) (- 2) (3) = - 6$ ✓.

Answer $k = - 1$

The sign of

k

is the commonest slip; always confirm with a numerical set.

Example 74 · Standard

Find

a^{2} (b - c) + b^{2} (c - a) + c^{2} (a - b)

for

a = 101, b = 102, c = 103

Solution, step by step

1The formula: $- (a - b) (b - c) (c - a)$ (from the previous example).
2The differences: $a - b = - 1$ , $b - c = - 1$ , $c - a = 2$ .
3Value $= - (- 1) (- 1) (2) = - 2$ . The direct way (check): $10201 (- 1) + 10404 (2) + 10609 (- 1) = - 10201 + 20808 - 10609 = - 2$ ✓.

Answer $- 2$

With big numbers the formula shrinks the work to three small differences.

Example 75 · Advanced

a^{3} (b - c) + b^{3} (c - a) + c^{3} (a - b) = k (a - b) (b - c) (c - a) (a + b + c)

find

k

, and give the value at

a = 1, b = 2, c = 4

Solution, step by step

1As before it is zero at $a = b$ , so $(a - b) (b - c) (c - a)$ is a factor. The expression has degree $4$ , the factors $3$ , so a symmetric factor of degree $1$ remains: $k (a + b + c)$ .
2 $(0, 1, 2)$ : the left side $= 0 + 1 \cdot 2 + 8 \cdot (- 1) = - 6$ ; $(a - b) (b - c) (c - a) = 2$ and $a + b + c = 3$ , so $k \cdot 2 \cdot 3 = - 6$ , $k = - 1$ .
3The value at $(1, 2, 4)$ : $- (- 1) (- 2) (3) (7) = - 42$ . Direct: $1 (- 2) + 8 (3) + 64 (- 1) = - 2 + 24 - 64 = - 42$ ✓.

Answer $k = - 1$ ;\ $- 42$

The degree count alone tells you that the extra factor

(a + b + c)

is there.

Example 76 · Advanced

Find

\frac{a ^{2}}{( a - b ) ( a - c )} + \frac{b ^{2}}{( b - c ) ( b - a )} + \frac{c ^{2}}{( c - a ) ( c - b )}

Solution, step by step

1Take the common bottom $D = (a - b) (b - c) (c - a)$ . $\frac{1}{( a - b ) ( a - c )} = - \frac{1}{( a - b ) ( c - a )} = - \frac{b - c}{D}$ , so the first term is $- \frac{a ^{2} ( b - c )}{D}$ ; likewise the second is $- \frac{b ^{2} ( c - a )}{D}$ and the third $- \frac{c ^{2} ( a - b )}{D}$ .
2The sum $= - \frac{a ^{2} ( b - c ) + b ^{2} ( c - a ) + c ^{2} ( a - b )}{D} = - \frac{- D}{D} = 1$ .
3Check at $(a, b, c) = (0, 1, 2)$ : the terms are $0$ , $r a c 1 (1 - 2) (1 - 0) = - 1$ and $r a c 4 (2 - 0) (2 - 1) = 2$ ; the sum is $1$ ✓.

Answer $1$

BasicTry it yourself first

Find

a^{2} (b - c) + b^{2} (c - a) + c^{2} (a - b)

for

a = 11, b = 12, c = 15

StandardTry it yourself first

Find

a^{2} (b - c) + b^{2} (c - a) + c^{2} (a - b)

for

a = 201, b = 203, c = 206

StandardTry it yourself first

Find

a (b^{2} - c^{2}) + b (c^{2} - a^{2}) + c (a^{2} - b^{2})

for

a = 3, b = 5, c = 9

StandardTry it yourself first

Find

a^{3} (b - c) + b^{3} (c - a) + c^{3} (a - b)

for

a = 2, b = 3, c = 7

AdvancedTry it yourself first

Find

(a + b + c)^{3} - a^{3} - b^{3} - c^{3}

for

a = 10, b = 13, c = 17

AdvancedTry it yourself first

Find

ab (a - b) + b c (b - c) + c a (c - a)

for

a = 4, b = 9, c = 13

BasicTry it yourself first

Which of these is cyclic but not symmetric?

StandardTry it yourself first

Which expression is not zero when

a = b

AdvancedTry it yourself first

What is the value of

\frac{a ^{2} ( b - c ) + b ^{2} ( c - a ) + c ^{2} ( a - b )}{( a - b ) ( b - c ) ( c - a )}

20. Least and greatest values

a square is never negative

A square being $\geq 0$ , the vertex of a parabola, $x + \frac{k}{x} \geq 2 k$ and the sum-product link.

The idea

Here value putting cannot find a least value (no single value tells you it beats all others); it can only test a claimed answer. Minimum and maximum questions have one source: a square is never negative. So when an expression can be written as “square $+$ number”, the least value occurs when the square is zero.

◭ Diagram

y = x^{2} - 6 x + 5 = (x - 3)^{2} - 4

. The square

(x - 3)^{2}

is least (

0

) at

x = 3

, so the least value of

y

- 4

. The vertex is that point.

∑ Formula

a x^{2} + b x + c = a (x + \frac{b}{2 a})^{2} + \frac{4 a c - b ^{2}}{4 a}, x + \frac{k}{x} \geq 2 k (x, k > 0)

Two main forms

$a, b, c$: quadratic coefficients; a least value for $a > 0$ , a greatest for $a < 0$
$x = - \frac{b}{2 a}$: where it occurs
$k$: a positive constant; equality at $x = k$

Sum and product: for two positive numbers with a fixed sum the product is greatest when they are equal (greatest product $\frac{S ^{2}}{4}$ for sum $S$ ). With a fixed product the sum is least when they are equal (least sum $2 P$ for product $P$ ). And if $x + y$ is fixed, $x^{2} + y^{2} \geq \frac{( x + y ) ^{2}}{2}$ .

CautionCaution:

x + \frac{k}{x} \geq 2 k

holds only for positive

x

; for negative

x

it is

\leq - 2 k

. And the least value is real only if the equality case (such as

x = k

) can be reached.

Steps

Bring the expression to “square $+$ constant” (complete the square) or spot $x + \frac{k}{x}$ .
Make the square zero: that gives the $x$ where the extreme value occurs.
Read the extreme value: the constant (a least value if $a > 0$ , a greatest value if $a < 0$ ).

Example 77 · Basic

Find the least value of

2 x^{2} - 12 x + 5

Solution, step by step

1 $a = 2 > 0$ , so there is a least value. Complete the square: $2 (x^{2} - 6 x) + 5 = 2 [(x - 3)^{2} - 9] + 5 = 2 (x - 3)^{2} - 13$ .
2 $(x - 3)^{2} \geq 0$ , so the expression is $\geq - 13$ , and $- 13$ is reached at $x = 3$ .
3Check with the formula: $\frac{4 a c - b ^{2}}{4 a} = \frac{40 - 144}{8} = - 13$ ✓.

Answer $- 13$

Example 78 · Standard

Find the greatest value of

7 + 6 x - x^{2}

Solution, step by step

1The coefficient of $x^{2}$ is $- 1 < 0$ , so there is a greatest value. $- (x^{2} - 6 x) + 7 = - [(x - 3)^{2} - 9] + 7 = 16 - (x - 3)^{2}$ .
2A non-negative $(x - 3)^{2}$ is being subtracted, so the expression is $\leq 16$ and $16$ is reached at $x = 3$ .

Answer $16$

A subtracted square makes a greatest value; an added square makes a least value.

Example 79 · Standard

For

x > 0

find the least value of

x + \frac{16}{x}

and the

x

at which it occurs. What happens for

x < 0

Solution, step by step

1Expanding $(x - \frac{4}{x})^{2} \geq 0$ gives $x - 8 + \frac{16}{x} \geq 0$ , that is $x + \frac{16}{x} \geq 8$ .
2Equality when $x = \frac{4}{x}$ , that is $x = 4$ . The least value is $8$ .
3For $x < 0$ , $x + \frac{16}{x} \leq - 8$ ( $- 8$ at $x = - 4$ ): there it can never be greater than $8$ .

Answer $8$ (at $x = 4$ )

Never use “

\geq 2 k

” without checking the sign of

x

Example 80 · Advanced

x + y = 10

, find the least value of

x^{2} + y^{2}

. Also, for positive

x, y

, what is the greatest value of

x y

Solution, step by step

1 $x^{2} + y^{2} = \frac{( x + y ) ^{2}}{2} + \frac{( x - y ) ^{2}}{2} = 50 + \frac{( x - y ) ^{2}}{2}$ (check: expanding the right gives $\frac{2 x ^{2} + 2 y ^{2}}{2}$ ).
2 $(x - y)^{2} \geq 0$ , so the least value is $50$ , when $x = y = 5$ .
3 $x y = \frac{( x + y ) ^{2} - ( x - y ) ^{2}}{4} = 25 - \frac{( x - y ) ^{2}}{4} \leq 25$ ; the greatest is $25$ , again at $x = y = 5$ .

Answer least $x^{2} + y^{2} = 50$ ; greatest $x y = 25$

One condition (

x = y

) gives the least of one and the greatest of the other.

BasicTry it yourself first

Find the least value of

x^{2} - 8 x + 3

StandardTry it yourself first

Find the least value of

3 x^{2} + 6 x + 7

StandardTry it yourself first

At which

x

does

2 x^{2} - 5 x + 1

take its least value?

AdvancedTry it yourself first

Find the greatest value of

- 2 x^{2} + 8 x + 5

AdvancedTry it yourself first

At which

x

does

- x^{2} + 4 x + 9

take its greatest value?

BasicTry it yourself first

For

x > 0

find the least value of

x + \frac{25}{x}

StandardTry it yourself first

For

x > 0

find the least value of

x + \frac{49}{x}

StandardTry it yourself first

Two positive numbers add up to

14

. Find the greatest possible product.

AdvancedTry it yourself first

Two positive numbers add up to

17

. Find the greatest possible product.

StandardTry it yourself first

x + y = 12

, find the least value of

x^{2} + y^{2}

AdvancedTry it yourself first

Find the least value of

(x - 3)^{2} + (y - 2)^{2} + 8

Summary, mixed practice, revision

All methods in one place

A table of all methods, exam-style mixed questions and revision cards.

Summary of all methods

Which method when

One page to pick the method as soon as you read a question.

Method	When	Key move
1-2 Componendo-dividendo	equal ratios, sum-difference asked	$\frac{a}{b} = \frac{c}{d} \Rightarrow \frac{a + b}{a - b} = \frac{c + d}{c - d}$
3 Square-root ladder	$\frac{A + B}{A - B}$	$\frac{x ^{2} + 1}{2 x} = \frac{A + B}{A - B}$
4 Cube form	$\frac{x ^{3} + 3 x}{3 x ^{2} + 1}$	$\frac{( x + 1 ) ^{3}}{( x - 1 ) ^{3}} = \frac{p + q}{p - q}$
5 Harmonic	$x = \frac{2 c d}{c + d}$	$\frac{x + c}{x - c} + \frac{x + d}{x - d} = 2$
6 Perfect squares	$(a + b) : ab$ , $\frac{a ^{2} + b ^{2}}{2 ab}$	twice the middle term
7 Limits	lone fraction, $a = b$	an equality and $A \neq = B$
8 The $k$ method	$\frac{a}{b} = \frac{c}{d} = \frac{e}{f}$	$= k$ , $\frac{\sum tops}{\sum bottoms} = k$
9 Homogeneous	a ratio, equal degrees	put the ratio numbers
10 Add/subtract 1	$\frac{k}{x + k}$ , $\frac{a}{b + c}$	$\frac{x}{x + k} = 1 - \frac{k}{x + k}$
11 $a + b + c = 0$	cubes, hidden zero sums	$a^{3} + b^{3} + c^{3} = 3 ab c$
12 $pq + q r + r p = 0$ , $ab c = 1$	$x^{2} - yz$ , $1 + a + ab$	$x^{2} - yz = x (x + y + z)$
13 Cycle	$x + \frac{1}{y} = 1$	$z + \frac{1}{x} = 1$ , $x yz = - 1$
14 Value putting	options, letters	small, different, legal numbers
15 Testing options	many options	numbers that separate; two tests
16 Values under a condition	a condition is given	choose all but one, last from the condition
17 Failure	coincidence, forbidden, half condition	check every condition, second set
18 Polynomials	remainder, factor, coefficient sum	$r = f (a)$ , $f (1)$ , $f (- 1)$
19 Symmetric/cyclic	differences like $(a - b)$	zero ⟹ factor
20 Least-greatest	quadratic, $x + \frac{k}{x}$	a square is $\geq 0$

Here: $a, b, c, d, x, y, z$ are the letters of the question, $A, B$ the quantities under roots, $k$ a common value or constant and $f (a)$ the value of the polynomial at $x = a$ .

First stepAfter reading ask: “Are there equal fractions?” Componendo-dividendo. “Is there a condition?” The move that simplifies it, or value putting. “Are there options?” Value putting, but with two tests.

Mixed practice — set 1

Basic to standard

All kinds mixed together; spot the method first, then solve.

The questions are not in method order. For each, first decide which move fits.

BasicTry it yourself first

\frac{a}{b} = \frac{11}{4}

, find

\frac{a + b}{a - b}

BasicTry it yourself first

\frac{x + y}{x - y} = \frac{8}{3}

, find

\frac{x}{y}

BasicTry it yourself first

x = \frac{10 + 6}{10 - 6}

, find

x + \frac{1}{x}

StandardTry it yourself first

\frac{x ^{3} + 3 x}{3 x ^{2} + 1} = \frac{559}{127}

, find

x

BasicTry it yourself first

x = 12

, find

\frac{x + 9}{x - 9} + \frac{x + 18}{x - 18}

BasicTry it yourself first

\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{5}{8}

and

b + d + f = 96

, find

a + c + e

StandardTry it yourself first

a : b = 3 : 5

and

b : c = 10 : 7

, find

\frac{a + c}{b}

StandardTry it yourself first

\frac{3}{x + 3} + \frac{6}{x + 6} + \dots + \frac{30}{x + 30} = 4

, find

\frac{x}{x + 3} + \frac{x}{x + 6} + \dots + \frac{x}{x + 30}

StandardTry it yourself first

Find the value of

3 1^{3} - 1 7^{3} - 1 4^{3}

StandardTry it yourself first

x y + yz + z x = 0

, what is

\frac{x}{x ^{2} - yz} + \frac{y}{y ^{2} - z x} + \frac{z}{z ^{2} - x y}

StandardTry it yourself first

x + \frac{1}{y} = 1

y + \frac{1}{z} = 1

and

x = 7

, find

z

StandardTry it yourself first

Which of these is equal to

(a + b)^{3} - 3 ab (a + b)

StandardTry it yourself first

a + b + c = 0

, find the value of

\frac{( b + c ) ( c + a ) ( a + b )}{ab c}

StandardTry it yourself first

Find the remainder when

x^{4} - 3 x^{3} + 2 x - 5

is divided by

(x - 2)

StandardTry it yourself first

Find the least value of

5 x^{2} - 20 x + 3

StandardTry it yourself first

Find

a^{2} (b - c) + b^{2} (c - a) + c^{2} (a - b)

for

a = 50, b = 51, c = 53

Mixed practice — set 2

Standard to advanced

Mixed questions from the hard end of the exam.

StandardTry it yourself first

\frac{3 p + 4 q}{3 p - 4 q} = \frac{9}{5}

, find

\frac{p}{q}

StandardTry it yourself first

\frac{3 x - 2 y}{x + 4 y} = \frac{3}{8}

, find

\frac{2 x + 5 y}{2 x - 5 y}

AdvancedTry it yourself first

x = \frac{12 + 3}{12 - 3}

, find

x^{2} + \frac{1}{x ^{2}}

AdvancedTry it yourself first

(a + b) : ab = 17 : 4

and

a > b > 0

, find

a : b

StandardTry it yourself first

x = \frac{120}{17}

(

= \frac{2 \cdot 15 \cdot 4}{4 ^{2} + 1}

), find

\frac{15 + x}{15 - x}

StandardTry it yourself first

\frac{a + b}{6} = \frac{b + c}{7} = \frac{c + a}{9}

, find

a : b : c

AdvancedTry it yourself first

\frac{p - 3 q}{p + 3 q} + \frac{q - 3 r}{q + 3 r} + \frac{r - 3 p}{r + 3 p} = - \frac{1}{2}

, find

\frac{p}{p + 3 q} + \frac{q}{q + 3 r} + \frac{r}{r + 3 p}

AdvancedTry it yourself first

\frac{a}{b + c} + \frac{b}{c + a} + \frac{c}{a + b} = 9

, find

(a + b + c) (\frac{1}{b + c} + \frac{1}{c + a} + \frac{1}{a + b})

AdvancedTry it yourself first

a + b + c = 0

and

a^{2} + b^{2} + c^{2} = 34

, find

a^{4} + b^{4} + c^{4}

StandardTry it yourself first

For

a = 3, b = 4, c = \frac{1}{12}

, find

\frac{1}{1 + a + ab} + \frac{1}{1 + b + b c} + \frac{1}{1 + c + c a}

AdvancedTry it yourself first

x + \frac{1}{y} = 1

and

y + \frac{1}{z} = 1

, then

x yz =

AdvancedTry it yourself first

a + b + c = 0

, what is

\frac{a ^{3} + b ^{3} + c ^{3}}{ab c}

AdvancedTry it yourself first

Which of these is equal to

(a + b + c)^{2} - (a - b - c)^{2}

AdvancedTry it yourself first

a + b = 7, ab = 12

, find the value of

a^{3} + b^{3}

AdvancedTry it yourself first

Find the sum of all the coefficients in the expansion of

(x^{2} - 3 x + 1)^{5}

AdvancedTry it yourself first

For

x > 0

find the least value of

x + \frac{36}{x}

StandardTry it yourself first

Find the value of

3 1^{3} - 1 7^{3} - 1 4^{3}

Revision — cards and traps

Flip and recall

Read the front, think, then flip.

Trap 1Trusting a single test. If several options look right at one set, test only the survivors with a second set.

Trap 2Equal values.

a = b = c = 1

erases terms like

(a - b)

and makes many options equal.

Trap 3Half a condition. With two conditions the set must meet both; assuming “all three fractions equal” can break a condition.

Trap 4Forbidden values. No zero denominator and keep limits such as

a > b

m > 1

; breaking a limit can cut even the right answer.

Trap 5Ratio numbers in a non-homogeneous expression. In

\frac{x + 1}{y + 1}

you cannot put the numbers of the ratio

2 : 3

Trap 6Cancelling $x + y + z$ . In

\frac{x}{y + z} = \frac{y}{z + x} = \frac{z}{x + y}

both

k = \frac{1}{2}

and

k = - 1

occur.

Trap 7“None of these”. Choose it only when each of A, B, C has been shown wrong by some legal set.