Algebraic Identities

SSC CGL · CHSL · CPO — method by method

5 Part81 worked examples255 practice questions

An identity is an equality that stays true for every value of its letters. $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ is an identity: put $a = 3, b = 4$ and both sides give 49; put $a = - 5, b = 0.5$ and both give 20.25. But $x + 3 = 5$ is an equation: it is true only for $x = 2$ .

∑ Formula

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

The three most important identities

$a, b$: any two numbers or expressions (for example $3 x$ and $2 y$ )

In exams identities do four jobs: (1) make big arithmetic small (like $9 7^{2}$ or $53 \times 47$ ); (2) find the value of an expression without solving for the letters (given $a + b$ and $ab$ , find $a^{3} + b^{3}$ ); (3) factorise and simplify; (4) answer questions on “least value”, “perfect squares” and “what follows if something is zero”. We learn them one method at a time — first two letters, then three, then questions with conditions, and last factorising and greatest/least values.

Habit: test itOne habit will help throughout: after learning any identity, test it with small values like

a = 2, b = 1

. It takes two minutes and catches a wrongly remembered formula (the last part has the full method).

Part 1: two letters

Squares · cubes · sum and product

Identities in $a$ and $b$ : squares, $a^{2} - b^{2}$ , sum and product, cubes, the make-a-cube move and the value of a polynomial.

Two-letter identities are the base of everything later. Learn them with understanding, not by rote.

Square of a sum and a difference

(a \pm b)^{2}

and quick squares

Why $(a + b)^{2}$ and $(a - b)^{2}$ look the way they do, expanding, spotting perfect squares and quick squares of numbers.

The idea

$(a + b)^{2}$ means $(a + b) \times (a + b)$ , the area of a square whose side is $a + b$ . Cut that square at $a$ : an $a \times a$ square at the top left, a small $b \times b$ square at the bottom right, and two equal rectangles in between, each $a \times b$ . The four pieces add up to the whole area. That is why the middle term is two times $ab$ , not one.

◭ Diagram

One colour = one piece: blue

a^{2}

, the two green rectangles

ab

and

ab

, saffron

b^{2}

For $(a - b)^{2}$ put $- b$ in place of $b$ . $a^{2}$ and $b^{2}$ stay as they are (because $(- b)^{2} = b^{2}$ ), but the middle $2 ab$ becomes $2 a (- b) = - 2 ab$ .

∑ Formula

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

The two square identities

$a, b$: any two numbers or expressions
$2 ab$: the middle term: twice the product of the two

Quick squares: break any number as “a round number $\pm$ a small gap”. For $98 = 100 - 2$ we get $9 8^{2} = 10000 - 400 + 4 = 9604$ . A number ending in 5: $(10 k + 5)^{2} = 100 k^{2} + 100 k + 25 = 100 k (k + 1) + 25$ . So multiply the tens digit $k$ by $k + 1$ and write 25 after it: for $3 5^{2}$ : $3 \times 4 = 12$ , answer 1225.

Spot itUse it when you see a squared bracket, a number close to 100, 1000 or 50, or the words “perfect square”.

Steps

Pick out the two terms and note the sign between them: $a$ and $\pm b$ .
Write the square of the first term (square its number too!).
Multiply the two terms, double it, and keep the sign that was inside the bracket.
Write the square of the second term, always with a $+$ .

Example 1 · Basic

Find

19 7^{2}

using a round number.

Solution, step by step

1 $197 = 200 - 3$ , so $a = 200$ and $b = 3$ ; the sign is minus.
2 $a^{2} = 40000$ , $2 ab = 2 \times 200 \times 3 = 1200$ , $b^{2} = 9$ .
3 $(a - b)^{2} = a^{2} - 2 ab + b^{2} = 40000 - 1200 + 9 = 38809$ .

Answer 38,809

Check by long multiplication:

197 \times 197 = 38809

✓.

Example 2 · Standard

Expand

(4 x - 5 y)^{2}

Solution, step by step

1 $a = 4 x$ , $b = 5 y$ , minus sign. Formula: $a^{2} - 2 ab + b^{2}$ .
2 $a^{2} = (4 x)^{2} = 16 x^{2}$ (the 4 is squared too), $b^{2} = (5 y)^{2} = 25 y^{2}$ .
3 $2 ab = 2 \times 4 x \times 5 y = 40 x y$ , taken with a minus sign.
4Answer: $16 x^{2} - 40 x y + 25 y^{2}$ .

Answer $16 x^{2} - 40 x y + 25 y^{2}$

Check: put

x = 1, y = 1

: left

(4 - 5)^{2} = 1

; right

16 - 40 + 25 = 1

✓.

Example 3 · Standard

9 x^{2} + k x + 16

is a perfect square. Give both possible values of

k

Solution, step by step

1A perfect square looks like $(a x \pm b)^{2} = a^{2} x^{2} \pm 2 ab x + b^{2}$ .
2 $a^{2} = 9$ gives $a = 3$ and $b^{2} = 16$ gives $b = 4$ .
3 $k = \pm 2 ab = \pm 2 \times 3 \times 4 = \pm 24$ . Both signs work: $(3 x + 4)^{2}$ and $(3 x - 4)^{2}$ .

Answer $k = 24$ / $k = - 24$

Trap: unless the question says “

k > 0

”, give both signs.

Example 4 · Advanced

Simplify:

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

Solution, step by step

1Expand both: $(2 x + 3)^{2} = 4 x^{2} + 12 x + 9$ and $(2 x - 3)^{2} = 4 x^{2} - 12 x + 9$ .
2Subtract: $4 x^{2}$ and $9$ cancel; $12 x - (- 12 x) = 24 x$ is left.
3Without expanding: $(a + b)^{2} - (a - b)^{2} = 4 ab = 4 \times 2 x \times 3 = 24 x$ .

Answer $24 x$

This

4 ab

result will keep returning.

Example 5 · Advanced

Find

8 5^{2}

and

2.3 5^{2}

without long multiplication.

Solution, step by step

1 $85 = 10 \times 8 + 5$ , so $k = 8$ . $k (k + 1) = 8 \times 9 = 72$ ; put 25 after it: $8 5^{2} = 7225$ .
2 $2.35 = \frac{235}{100}$ , so first take $23 5^{2}$ : $k = 23$ , $23 \times 24 = 552$ , so $23 5^{2} = 55225$ .
3Two decimal places squared give four: $2.3 5^{2} = 5.5225$ .

Answer $7225$ ; $5.5225$

(2.35)^{2} = \frac{23 5 ^{2}}{10 0 ^{2}}

— count: the point moves

2 + 2 = 4

places.

Practice

BasicTry it yourself first

Find the value of

10 3^{2}

BasicTry it yourself first

Find the value of

4 8^{2}

StandardTry it yourself first

Find the value of

1, 00 4^{2}

StandardTry it yourself first

Find the value of

49 7^{2}

BasicTry it yourself first

The square of a number ending in 5:

6 5^{2}

StandardTry it yourself first

The square of a number ending in 5:

14 5^{2}

BasicTry it yourself first

Expand

(2 x + 3 y)^{2}

BasicTry it yourself first

Expand

(5 x - y)^{2}

StandardTry it yourself first

Expand

(7 x - 4 y)^{2}

StandardTry it yourself first

25 x^{2} + k x + 36

is a perfect square and

k > 0

. Find

k

StandardTry it yourself first

What last term makes

36 x^{2} - 84 x + ?

a perfect square?

AdvancedTry it yourself first

Simplify:

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

AdvancedTry it yourself first

y^{2} + 16 y = 20

, find the value of

(y + 8)^{2}

AdvancedTry it yourself first

81 x^{2} + k x + 4

is a perfect square and

k > 0

. Find

k

Difference of two squares

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

Factorising a difference of squares, products of numbers at equal distances and extensions like $a^{4} - b^{4}$ .

The idea

Cut a corner off an $a \times a$ square: a small $b \times b$ square. What is left is an L-shape of area $a^{2} - b^{2}$ . Cut the L into two rectangles and turn the upper one round to the right: it becomes one long rectangle of length $a + b$ and width $a - b$ . The areas are equal, so $a^{2} - b^{2} = (a + b) (a - b)$ .

◭ Diagram

The saffron piece turns; the blue piece stays; the dashed square is the corner that was cut away

∑ Formula

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

Difference of squares

$a, b$: any two numbers or expressions
$a + b, a - b$: their sum and their difference

It can be read both ways. Left to right: turn a difference into a product (factorising), like $x^{2} - 49 = (x + 7) (x - 7)$ . Right to left: the product of two numbers at equal distances from a middle number: $47 \times 53 = (50 - 3) (50 + 3) = 2500 - 9 = 2347$ . Two more uses: $n^{2} - (n - 1)^{2} = 2 n - 1$ and $(a + b)^{2} - (a - b)^{2} = 4 ab$ .

∑ Formula

a^{4} - b^{4} = (a^{2} + b^{2}) (a + b) (a - b), (a - b) (a + b) (a^{2} + b^{2}) (a^{4} + b^{4}) = a^{8} - b^{8}

Applied repeatedly

$a^{2} + b^{2}$: this does not factorise further (over real numbers)

Spot itSpot it: two terms, both perfect squares, with a minus between. There is no such factorisation for

a^{2} + b^{2}

— a plus sign does not fit this formula.

Steps

Take the square root of each term (of the numbers too): these are $a$ and $b$ .
Write $(a + b)$ and $(a - b)$ .
For a numeric product take the middle number as $a$ and the distance as $b$ .
Expand once to check.

Example 6 · Basic

Find

63 \times 57

with the identity.

Solution, step by step

1The number between 63 and 57 is 60; both are 3 away: $63 = 60 + 3$ , $57 = 60 - 3$ .
2 $(60 + 3) (60 - 3) = 6 0^{2} - 3^{2}$ .
3 $3600 - 9 = 3591$ .

Answer 3,591

Check:

63 \times 57 = 3591

✓ (long multiplication agrees).

Example 7 · Standard

Find

\frac{1 , 00 4 ^{2} - 99 6 ^{2}}{8}

Solution, step by step

1The numerator is a difference of squares: $(1004 + 996) (1004 - 996)$ .
2 $1004 + 996 = 2000$ and $1004 - 996 = 8$ .
3 $\frac{2000 \times 8}{8} = 2000$ . The 8 in the denominator cancelled the 8 in the numerator.

Answer 2,000

Squaring first wastes time and invites errors.

Example 8 · Standard

Factorise

49 x^{2} - 81 y^{2}

and

x^{4} - 16

Solution, step by step

1 $49 x^{2} = (7 x)^{2}$ and $81 y^{2} = (9 y)^{2}$ , so $49 x^{2} - 81 y^{2} = (7 x + 9 y) (7 x - 9 y)$ .
2 $x^{4} - 16 = (x^{2})^{2} - 4^{2} = (x^{2} + 4) (x^{2} - 4)$ .
3 $x^{2} - 4$ is again a difference of squares: $(x + 2) (x - 2)$ .
4Complete: $x^{4} - 16 = (x^{2} + 4) (x + 2) (x - 2)$ . $x^{2} + 4$ does not break further.

Answer $(7 x + 9 y) (7 x - 9 y)$ ; $(x^{2} + 4) (x + 2) (x - 2)$

Keep breaking while some factor is still a difference of squares.

Example 9 · Advanced

Find

(5 - 1) (5 + 1) (5^{2} + 1) (5^{4} + 1)

Solution, step by step

1The first two factors: $(5 - 1) (5 + 1) = 5^{2} - 1$ .
2Now $(5^{2} - 1) (5^{2} + 1) = 5^{4} - 1$ (the same identity again, with $a = 5^{2}$ ).
3Now $(5^{4} - 1) (5^{4} + 1) = 5^{8} - 1$ .
4 $5^{8} = 390625$ , so the answer is $390624$ .

Answer 3,90,624

In big-number questions look for “the power doubles at each step”.

Practice

BasicTry it yourself first

Find

37 \times 43

using the identity.

BasicTry it yourself first

Find

70 \times 80

using the identity.

BasicTry it yourself first

What is the value of

21 4^{2} - 21 3^{2}

StandardTry it yourself first

Find

993 \times 1, 007

using the identity.

StandardTry it yourself first

Find the value of

\frac{6. 8 ^{2} - 3. 2 ^{2}}{6.8 + 3.2}

StandardTry it yourself first

Find the value of

\frac{9. 5 ^{2} - 4. 5 ^{2}}{9.5 + 4.5}

StandardTry it yourself first

Factorise

36 x^{2} - 25 y^{2}

StandardTry it yourself first

Factorise

121 x^{2} - 16 y^{2}

StandardTry it yourself first

What is the value of

1, 50 0^{2} - 1, 49 9^{2}

StandardTry it yourself first

Find

(7.3 + 2.7)^{2} - (7.3 - 2.7)^{2}

AdvancedTry it yourself first

a + b = 7

a - b = 3

and

a^{2} + b^{2} = 29

, find

a^{4} - b^{4}

AdvancedTry it yourself first

a + b = 11

a - b = 3

and

a^{2} + b^{2} = 65

, find

a^{4} - b^{4}

AdvancedTry it yourself first

Find

(3 - 1) (3 + 1) (3^{2} + 1) (3^{4} + 1)

AdvancedTry it yourself first

Find

(2 - 1) (2 + 1) (2^{2} + 1) (2^{4} + 1)

Values from a sum and a product

When

a + b

and

ab

are given

Finding $a^{2} + b^{2}$ , $a^{4} + b^{4}$ , $a - b$ and fractions when the sum or difference and the product are given.

The idea

Many questions do not give $a$ and $b$ separately; they give only the sum $a + b$ (or difference) and the product $ab$ . We do not need to find $a$ and $b$ (that would mean solving a quadratic). The square identity lets us write $a^{2} + b^{2}$ directly with $a + b$ and $ab$ . Such expressions are symmetric: swapping $a$ and $b$ does not change them, so they can always be built from the sum and the product.

◭ Diagram

The big square has side

a + b

; the four green rectangles are

ab

each; the pink square in the middle is

(a - b)^{2}

∑ Formula

a^{2} + b^{2} = (a + b)^{2} - 2 ab, (a - b)^{2} = (a + b)^{2} - 4 ab, a^{2} + b^{2} = (a - b)^{2} + 2 ab

The basic links between two letters

$a + b$: their sum (given)
$ab$: their product (given)
$a - b$: their difference

To find	Formula ( $S = a + b, P = ab$ )
$a^{2} + b^{2}$	$S^{2} - 2 P$
$(a - b)^{2}$	$S^{2} - 4 P$
$a^{4} + b^{4}$	$(S^{2} - 2 P)^{2} - 2 P^{2}$
$\frac{1}{a} + \frac{1}{b}$	$\frac{S}{P}$
$\frac{a}{b} + \frac{b}{a}$	$\frac{S ^{2} - 2 P}{P}$

Here $S = a + b$ (the sum) and $P = ab$ (the product). Two more points: (1) when you get the value of $(a + b)^{2}$ or $(a - b)^{2}$ and need $a + b$ , the square root has two signs unless the question says $a, b$ are positive; (2) the numbers in such a question must fit real $a, b$ (in exams they do).

Spot itSpot it: “

a + b = \dots

and

ab = \dots

”, “

a - b = \dots

and

ab = \dots

” or “

a^{2} + b^{2} = \dots

” is given and something from the list is asked.

Steps

Write what is given ( $S$ and $P$ ) and what is asked.
Write the asked expression using $S$ and $P$ (from the table or the square identity).
Put in the values and calculate.
If you need $a + b$ or $a - b$ , take the square root and watch the sign.

Example 10 · Basic

a + b = 9

and

ab = 14

, find

a^{2} + b^{2}

and

(a - b)^{2}

Solution, step by step

1 $a^{2} + b^{2} = (a + b)^{2} - 2 ab = 81 - 28 = 53$ .
2 $(a - b)^{2} = (a + b)^{2} - 4 ab = 81 - 56 = 25$ .
3Check: $a = 7, b = 2$ also fit the data: $a^{2} + b^{2} = 49 + 4 = 53$ ✓.

Answer $53$ ; $25$

Example 11 · Standard

a - b = 4

and

ab = 21

, find

a^{2} + b^{2}

and

a + b

Solution, step by step

1 $a^{2} + b^{2} = (a - b)^{2} + 2 ab = 16 + 42 = 58$ .
2 $(a + b)^{2} = (a - b)^{2} + 4 ab = 16 + 84 = 100$ .
3Square root: $a + b = \pm 10$ . Both signs are right: $(7, 3)$ gives $a + b = 10$ and $(- 3, - 7)$ gives $- 10$ ; both pairs have $a - b = 4$ and $ab = 21$ .

Answer $58$ ; $a + b = \pm 10$

If nothing says

a, b

are positive, write ±.

Example 12 · Standard

a > b > 0

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

and

ab = 15

. Find

a + b

and

a - b

Solution, step by step

1 $(a + b)^{2} = a^{2} + b^{2} + 2 ab = 34 + 30 = 64$ , and $a + b > 0$ , so $a + b = 8$ .
2 $(a - b)^{2} = a^{2} + b^{2} - 2 ab = 34 - 30 = 4$ , and $a > b$ , so $a - b = 2$ .
3(If you like, you can then get $a = 5, b = 3$ .)

Answer $8$ ; $2$

Example 13 · Advanced

a + b = 6

and

ab = 7

, find

a^{4} + b^{4}

Solution, step by step

1First $a^{2} + b^{2} = 36 - 14 = 22$ .
2 $a^{4} + b^{4} = (a^{2} + b^{2})^{2} - 2 a^{2} b^{2}$ , because $(a^{2} + b^{2})^{2} = a^{4} + b^{4} + 2 a^{2} b^{2}$ .
3 $a^{2} b^{2} = 49$ , so $2 2^{2} - 2 \times 49 = 484 - 98 = 386$ .

Answer $386$

Here

a, b

are not whole numbers (

a, b = 3 \pm 2

), yet we got the answer without finding them — that is the strength of the method.

Example 14 · Advanced

a + b = 10

ab = 16

, find

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

and

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

Solution, step by step

1 $\frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab} = \frac{10}{16} = \frac{5}{8}$ .
2 $\frac{a}{b} + \frac{b}{a} = \frac{a ^{2} + b ^{2}}{ab}$ , and $a^{2} + b^{2} = 100 - 32 = 68$ .
3 $\frac{68}{16} = \frac{17}{4}$ . Check with $a = 8, b = 2$ : $4 + \frac{1}{4} = \frac{17}{4}$ ✓.

Answer $\frac{5}{8}$ ; $\frac{17}{4}$

Practice

BasicTry it yourself first

a + b = 8

and

ab = 15

, find

a^{2} + b^{2}

BasicTry it yourself first

a + b = 13

and

ab = 36

, find

a^{2} + b^{2}

BasicTry it yourself first

a - b = 5

and

ab = 24

, find

a^{2} + b^{2}

StandardTry it yourself first

a - b = 7

and

ab = 44

, find

a^{2} + b^{2}

StandardTry it yourself first

For

a > b > 0

a^{2} + b^{2} = 61

and

ab = 30

, find

a + b

StandardTry it yourself first

For

a > b > 0

a^{2} + b^{2} = 97

and

ab = 36

, find

a - b

StandardTry it yourself first

a + b = 7

and

ab = 10

, find

a^{4} + b^{4}

AdvancedTry it yourself first

a + b = 10

and

ab = 21

, find

a^{4} + b^{4}

StandardTry it yourself first

a + b = 9

and

ab = 18

, find

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

AdvancedTry it yourself first

a + b = 13

and

ab = 36

, find

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

AdvancedTry it yourself first

For

a > b > 0

a^{2} + b^{2} = 169

and

ab = 60

, find

a + b

AdvancedTry it yourself first

a + b = 15

and

ab = 36

, find

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

Cube of a sum and a difference

(a \pm b)^{3}

and its link to

a^{3} \pm b^{3}

The eight pieces of $(a + b)^{3}$ , expanding, and getting $a^{3} + b^{3}$ from $a + b$ and $ab$ .

The idea

$(a + b)^{3}$ is the volume of a cube with side $a + b$ . Cut it at $a$ in every direction and you get eight pieces: one big cube $a^{3}$ , one small cube $b^{3}$ , three flat slabs $a \times a \times b$ (each $a^{2} b$ ) and three long rods $a \times b \times b$ (each $a b^{2}$ ). Add the volumes: $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ . That is where the 3 comes from.

◭ Diagram

Colour = type of block: blue

a^{3}

, three green slabs

a^{2} b

, three saffron rods

a b^{2}

, pink

b^{3}

∑ Formula

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} = a^{3} + b^{3} + 3 ab (a + b)

Cube of a sum

$a, b$: any two numbers or expressions
$3 ab (a + b)$: the two middle terms grouped: $3 a^{2} b + 3 a b^{2}$

∑ Formula

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

Cube of a difference (put

- b

for

b

)

$a, b$: any two numbers or expressions

Putting $- b$ for $b$ changes the signs of $b^{3}$ and $3 a^{2} b$ (as $(- b)^{3} = - b^{3}$ ) but not of $3 a b^{2}$ (as $(- b)^{2} = b^{2}$ ). So in the cube of a difference the signs alternate: $+, -, +, -$ . Both formulas can be turned round, and that is what exams use:

∑ Formula

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

Sum/difference of cubes from the sum/difference and the product

$a + b, a - b$: the given sum or difference
$ab$: the given product

Spot itSpot it: a cubed bracket; or

a + b

and

ab

given and

a^{3} + b^{3}

asked; or an expression hiding the shape

a^{3} + b^{3} + 3 ab (\dots)

Steps

Identify $a$ and $b$ (with the sign).
Find the four terms separately: $a^{3}, 3 a^{2} b, 3 a b^{2}, b^{3}$ — cube the numbers too.
In a cube of a difference put the minus on the second and the last term.
In sum/difference and product questions use the turned-round form.

Example 15 · Basic

Expand

(2 x + 3)^{3}

Solution, step by step

1 $a = 2 x$ , $b = 3$ . $a^{3} = 8 x^{3}$ (2 cubed is 8).
2 $3 a^{2} b = 3 \times 4 x^{2} \times 3 = 36 x^{2}$ , $3 a b^{2} = 3 \times 2 x \times 9 = 54 x$ , $b^{3} = 27$ .
3 $(2 x + 3)^{3} = 8 x^{3} + 36 x^{2} + 54 x + 27$ .

Answer $8 x^{3} + 36 x^{2} + 54 x + 27$

Check:

x = 1

5^{3} = 125

and

8 + 36 + 54 + 27 = 125

✓.

Example 16 · Standard

Expand

(3 p - 2 q)^{3}

Solution, step by step

1 $a = 3 p$ , $b = 2 q$ ; the signs will be $+, -, +, -$ .
2 $a^{3} = 27 p^{3}$ ; $3 a^{2} b = 3 \times 9 p^{2} \times 2 q = 54 p^{2} q$ ; $3 a b^{2} = 3 \times 3 p \times 4 q^{2} = 36 p q^{2}$ ; $b^{3} = 8 q^{3}$ .
3 $27 p^{3} - 54 p^{2} q + 36 p q^{2} - 8 q^{3}$ .

Answer $27 p^{3} - 54 p^{2} q + 36 p q^{2} - 8 q^{3}$

Check:

p = q = 1

(3 - 2)^{3} = 1

and

27 - 54 + 36 - 8 = 1

✓.

Example 17 · Standard

a + b = 5

and

ab = 6

, find

a^{3} + b^{3}

Solution, step by step

1 $a^{3} + b^{3} = (a + b)^{3} - 3 ab (a + b)$ — we do not need $a, b$ separately.
2 $(a + b)^{3} = 125$ and $3 ab (a + b) = 3 \times 6 \times 5 = 90$ .
3 $125 - 90 = 35$ . Check: $a = 3, b = 2$ : $27 + 8 = 35$ ✓.

Answer 35

Example 18 · Advanced

a - b = 3

and

ab = 10

, find

a^{3} - b^{3}

Solution, step by step

1 $(a - b)^{3} = a^{3} - b^{3} - 3 ab (a - b)$ , so $a^{3} - b^{3} = (a - b)^{3} + 3 ab (a - b)$ .
2 $27 + 3 \times 10 \times 3 = 27 + 90 = 117$ .
3Check: $a = 5, b = 2$ : $a - b = 3$ , $ab = 10$ , $125 - 8 = 117$ ✓.

Answer 117

Trap: in the difference formula the middle part is added, not subtracted.

Example 19 · Advanced

x + y = 1

, what is the value of

x^{3} + y^{3} + 3 x y

Solution, step by step

1 $(x + y)^{3} = x^{3} + y^{3} + 3 x y (x + y)$ .
2Putting $x + y = 1$ : $1^{3} = x^{3} + y^{3} + 3 x y \times 1$ .
3So $x^{3} + y^{3} + 3 x y = 1$ . We knew nothing else about $x$ and $y$ , yet we got the value.

Answer 1

A classic exam move:

x + y

is a number and

3 x y

carries a multiplier equal to it.

Practice

BasicTry it yourself first

Expand

(2 x + y)^{3}

BasicTry it yourself first

Expand

(x - 4 y)^{3}

StandardTry it yourself first

Expand

(3 x + 2 y)^{3}

StandardTry it yourself first

Expand

(4 x - 3 y)^{3}

BasicTry it yourself first

Find the value of

1 2^{3}

StandardTry it yourself first

Find the value of

1 7^{3}

StandardTry it yourself first

a + b = 7

and

ab = 12

, find

a^{3} + b^{3}

StandardTry it yourself first

a + b = 9

and

ab = 14

, find

a^{3} + b^{3}

StandardTry it yourself first

a - b = 4

and

ab = 12

, find

a^{3} - b^{3}

AdvancedTry it yourself first

a - b = 5

and

ab = 36

, find

a^{3} - b^{3}

AdvancedTry it yourself first

x + y = 4

, find

x^{3} + y^{3} + 12 x y

AdvancedTry it yourself first

x + y = 7

, find

x^{3} + y^{3} + 21 x y

AdvancedTry it yourself first

a + b = 5

and

a^{3} + b^{3} = 65

, find

ab

AdvancedTry it yourself first

a + b = 9

and

a^{3} + b^{3} = 243

, find

ab

Sum and difference of cubes

Factors of

a^{3} \pm b^{3}

Factors of $a^{3} + b^{3}$ and $a^{3} - b^{3}$ , when they cancel, and up to $a^{6} - b^{6}$ .

The idea

Take $a^{3} + b^{3} = (a + b)^{3} - 3 ab (a + b)$ from the previous method and pull out $(a + b)$ : $a^{3} + b^{3} = (a + b) [(a + b)^{2} - 3 ab] = (a + b) (a^{2} + 2 ab + b^{2} - 3 ab) = (a + b) (a^{2} - ab + b^{2})$ . In the same way $a^{3} - b^{3} = (a - b) [(a - b)^{2} + 3 ab] = (a - b) (a^{2} + ab + b^{2})$ . So a sum or difference of cubes always splits into two factors: a simple $a \pm b$ and a three-term one.

∑ Formula

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

Sum and difference of cubes

$a, b$: any two numbers or expressions
$a^{2} \mp ab + b^{2}$: the three-term factor

A way to remember the signs: the first bracket keeps the sign of the question. In the second bracket the middle sign is the opposite and the last term is always $+ b^{2}$ . To test, put $a = 2, b = 1$ : $a^{3} + b^{3} = 9$ and $(a + b) (a^{2} - ab + b^{2}) = 3 \times 3 = 9$ ✓.

∑ Formula

a^{6} - b^{6} = (a^{3} - b^{3}) (a^{3} + b^{3}) = (a - b) (a + b) (a^{2} + ab + b^{2}) (a^{2} - ab + b^{2})

Sixth power: both formulas together

$a^{6} - b^{6}$: first a difference of squares $(a^{3})^{2} - (b^{3})^{2}$ , then each cube is split

Spot itSpot it: two terms that are both perfect cubes (8, 27, 64, 125, 216 …), or a shape like

a^{2} \mp ab + b^{2}

in a numerator or denominator.

Steps

Take the cube root of each term: these are $a$ and $b$ (of the number too).
Write the first bracket $(a \pm b)$ .
Write the second bracket $a^{2} \mp ab + b^{2}$ : opposite middle sign, last term $+$ .
In a fraction, cancel the matching bracket.

Example 20 · Basic

Factorise

27 x^{3} + 8 y^{3}

Solution, step by step

1 $27 x^{3} = (3 x)^{3}$ and $8 y^{3} = (2 y)^{3}$ , so $a = 3 x$ , $b = 2 y$ ; the sign is $+$ .
2First bracket $(3 x + 2 y)$ . Second: $a^{2} - ab + b^{2} = 9 x^{2} - 6 x y + 4 y^{2}$ .
3 $27 x^{3} + 8 y^{3} = (3 x + 2 y) (9 x^{2} - 6 x y + 4 y^{2})$ .

Answer $(3 x + 2 y) (9 x^{2} - 6 x y + 4 y^{2})$

Check

x = y = 1

35

and

5 \times 7 = 35

✓.

Example 21 · Standard

Find

\frac{8. 4 ^{3} + 1. 6 ^{3}}{8. 4 ^{2} - 8.4 \times 1.6 + 1. 6 ^{2}}

Solution, step by step

1The numerator is $a^{3} + b^{3}$ ( $a = 8.4$ , $b = 1.6$ ) and the denominator is exactly $a^{2} - ab + b^{2}$ .
2 $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$ , so the fraction equals $a + b$ .
3 $8.4 + 1.6 = 10$ .

Answer 10

Example 22 · Standard

Factorise

x^{3} - 125

and simplify

\frac{x ^{3} - 125}{x - 5}

Solution, step by step

1 $125 = 5^{3}$ , so $a = x$ , $b = 5$ ; the sign is $-$ .
2 $x^{3} - 5^{3} = (x - 5) (x^{2} + 5 x + 25)$ — the middle sign in the second bracket is $+$ .
3In the fraction $(x - 5)$ cancels: the answer is $x^{2} + 5 x + 25$ ( $x \neq = 5$ ).

Answer $(x - 5) (x^{2} + 5 x + 25)$ ; $x^{2} + 5 x + 25$

x = 5

the denominator is zero, so that value is excluded.

Example 23 · Advanced

Factorise

a^{6} - b^{6}

completely.

Solution, step by step

1 $a^{6} - b^{6} = (a^{3})^{2} - (b^{3})^{2} = (a^{3} + b^{3}) (a^{3} - b^{3})$ (difference of squares).
2 $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$ and $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ .
3 $a^{6} - b^{6} = (a - b) (a + b) (a^{2} + ab + b^{2}) (a^{2} - ab + b^{2})$ .

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

Using the square formula first gives all four factors; using the cube formula first would leave

a^{2} - b^{2}

unfinished.

Example 24 · Advanced

a - b = 2

, find

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

Solution, step by step

1 $(a - b)^{3} = a^{3} - b^{3} - 3 ab (a - b)$ .
2Put $a - b = 2$ : $2^{3} = a^{3} - b^{3} - 3 ab \times 2 = a^{3} - b^{3} - 6 ab$ .
3So the value is $8$ .

Answer 8

Practice

BasicTry it yourself first

Factorise

8 x^{3} + y^{3}

BasicTry it yourself first

Factorise

27 x^{3} - y^{3}

StandardTry it yourself first

Factorise

64 x^{3} + 27 y^{3}

StandardTry it yourself first

Factorise

125 x^{3} - 8 y^{3}

StandardTry it yourself first

Find

\frac{7. 6 ^{3} + 2. 4 ^{3}}{7. 6 ^{2} - 7.6 \times 2.4 + 2. 4 ^{2}}

StandardTry it yourself first

Find

\frac{9. 3 ^{3} - 1. 7 ^{3}}{9. 3 ^{2} + 9.3 \times 1.7 + 1. 7 ^{2}}

StandardTry it yourself first

Simplify:

\frac{x ^{3} + 27}{x + 3}

StandardTry it yourself first

Simplify:

\frac{x ^{3} - 64}{x - 4}

StandardTry it yourself first

a + b = 8

and

a^{2} - ab + b^{2} = 19

, find

a^{3} + b^{3}

AdvancedTry it yourself first

a - b = 5

and

a^{2} + ab + b^{2} = 67

, find

a^{3} - b^{3}

StandardTry it yourself first

Find

\frac{1 0 ^{3} + 7 ^{3}}{17}

AdvancedTry it yourself first

Find

\frac{1 2 ^{3} + 5 ^{3}}{17}

AdvancedTry it yourself first

a - b = 3

, find

a^{3} - b^{3} - 9 ab

AdvancedTry it yourself first

a - b = 5

, find

a^{3} - b^{3} - 15 ab

The make-a-cube move

A cubic expression at a big number

Turning an expression into $(x \pm m)^{3} \mp m^{3}$ to get its value at a big $x$ , and cubes of numbers near 100.

The idea

Suppose the expression is $x^{3} + 9 x^{2} + 27 x$ and $x = 97$ . Putting it in directly means three big multiplications. But notice $(x + 3)^{3} = x^{3} + 9 x^{2} + 27 x + 27$ . Our expression is just missing the last term 27. So $x^{3} + 9 x^{2} + 27 x = (x + 3)^{3} - 27$ , and $x + 3 = 100$ is a round number: $10 0^{3} - 27 = 999973$ . That is the move: complete the cube that is incomplete, then take away what you added.

∑ Formula

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

The incomplete cube

$x$: the given value (for example 97)
$m$: one third of the coefficient of $x^{2}$ (may be negative)
$x + m$: this is a round number; the question is built for it

How to recognise it: the coefficient of $x^{2}$ is $3 m$ and that of $x$ is $3 m^{2}$ . For $x^{3} - 12 x^{2} + 48 x$ : $3 m = - 12 \Rightarrow m = - 4$ and $3 m^{2} = 48$ ✓. Then the expression is $(x - 4)^{3} + 64$ (since $(- 4)^{3} = - 64$ , add $64$ to cancel it). Such questions are often written as “ $x \times (\dots)$ ”, which hides the $x^{3}$ .

Spot itSpot it:

x

is a big number (97, 52, 198 …); the expression has

x^{3}, x^{2}, x

, and the coefficients of

x^{2}

and

x

follow “

3 m

and

3 m^{2}

”.

Steps

Divide the coefficient of $x^{2}$ by 3 to get $m$ (with its sign).
Check that the coefficient of $x$ is $3 m^{2}$ .
Write the expression as $(x + m)^{3} - m^{3}$ (adjust if there was a last term).
Put in the value of $x + m$ , cube it, and subtract $m^{3}$ .

Example 25 · Basic

For

x = 52

, find

x (x^{2} - 6 x + 12)

Solution, step by step

1Multiply out: $x^{3} - 6 x^{2} + 12 x$ . Here $3 m = - 6$ , so $m = - 2$ ; and $3 m^{2} = 12$ ✓.
2 $(x - 2)^{3} = x^{3} - 6 x^{2} + 12 x - 8$ . Our expression has no $- 8$ , so $x^{3} - 6 x^{2} + 12 x = (x - 2)^{3} + 8$ .
3 $x - 2 = 50$ , so $5 0^{3} + 8 = 125000 + 8 = 125008$ .

Answer 1,25,008

Check (with a small

x

x = 3

27 - 54 + 36 = 9

and

(3 - 2)^{3} + 8 = 9

✓.

Example 26 · Standard

For

x = 97

, find

x^{3} + 9 x^{2} + 27 x

Solution, step by step

1 $3 m = 9 \Rightarrow m = 3$ and $3 m^{2} = 27$ ✓.
2 $(x + 3)^{3} = x^{3} + 9 x^{2} + 27 x + 27$ , so the expression $= (x + 3)^{3} - 27$ .
3 $x + 3 = 100$ : $10 0^{3} - 27 = 1000000 - 27 = 999973$ .

Answer 9,99,973

Example 27 · Standard

Find

10 8^{3}

from the four pieces of

(100 + 8)^{3}

Solution, step by step

1 $a = 100$ , $b = 8$ : $a^{3} = 1000000$ .
2 $3 a^{2} b = 3 \times 10000 \times 8 = 240000$ , $3 a b^{2} = 3 \times 100 \times 64 = 19200$ , $b^{3} = 512$ .
3 $1000000 + 240000 + 19200 + 512 = 1259712$ .

Answer 12,59,712

This is the “cubes near 100” method; each piece is easy to find alone.

Example 28 · Advanced

For

x = 45

, find

x (x^{2} + 15 x + 75)

Solution, step by step

1Expanded: $x^{3} + 15 x^{2} + 75 x$ . $3 m = 15 \Rightarrow m = 5$ ; $3 m^{2} = 75$ ✓.
2 $(x + 5)^{3} = x^{3} + 15 x^{2} + 75 x + 125$ , so the expression $= (x + 5)^{3} - 125$ .
3 $x + 5 = 50$ : $125000 - 125 = 124875$ .

Answer 1,24,875

Example 29 · Advanced

For

x = 13

, find

x^{3} - 9 x^{2} + 27 x - 27

Solution, step by step

1All four terms are present: $3 m = - 9 \Rightarrow m = - 3$ ; $3 m^{2} = 27$ ✓; and $m^{3} = - 27$ ✓ — it is a complete cube.
2The expression $= (x - 3)^{3}$ .
3 $x - 3 = 10$ , so the value is $1 0^{3} = 1000$ .

Answer 1,000

When all four terms are there, no “

- m^{3}

” adjustment is needed.

Practice

BasicTry it yourself first

For

x = 48

, find the value of

x^{3} + 6 x^{2} + 12 x

BasicTry it yourself first

For

x = 197

, find the value of

x^{3} + 9 x^{2} + 27 x

StandardTry it yourself first

For

x = 101

, find the value of

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

StandardTry it yourself first

For

x = 96

, find the value of

x^{3} + 12 x^{2} + 48 x

StandardTry it yourself first

For

x = 303

, find the value of

x^{3} - 9 x^{2} + 27 x

StandardTry it yourself first

For

x = 95

, find

x^{3} + 15 x^{2} + 75 x + 125

StandardTry it yourself first

For

x = 104

, find

x^{3} - 12 x^{2} + 48 x - 64

BasicTry it yourself first

Find

10 6^{3}

by treating it as

(100 + 6)^{3}

StandardTry it yourself first

Find

10 9^{3}

by treating it as

(100 + 9)^{3}

StandardTry it yourself first

Find

9 6^{3}

by treating it as

(100 - 4)^{3}

AdvancedTry it yourself first

For

x = 94

, find the value of

x^{3} + 18 x^{2} + 108 x

AdvancedTry it yourself first

For

x = 205

, find the value of

x^{3} - 15 x^{2} + 75 x

AdvancedTry it yourself first

Find

9 3^{3}

by treating it as

(100 - 7)^{3}

AdvancedTry it yourself first

For

x = 202

, find

x^{3} - 6 x^{2} + 12 x - 8

Value of a polynomial: write the coefficient with $x$

Terms cancel in pairs

When the same coefficient repeats and equals $x \pm 1$ , getting the value of a polynomial without big powers.

The idea

Let $x = 9$ and the polynomial be $x^{4} - 10 x^{3} + 10 x^{2} - 10 x + 10$ . The number 10 repeats, and $10 = x + 1$ . Rewrite: $x^{4} - (x + 1) x^{3} + (x + 1) x^{2} - (x + 1) x + (x + 1)$ . Now $x^{4} - (x + 1) x^{3} = x^{4} - x^{4} - x^{3} = - x^{3}$ . Then $- x^{3} + (x + 1) x^{2} = x^{2}$ , then $x^{2} - (x + 1) x = - x$ , then $- x + (x + 1) = 1$ . All the powers cancelled; the answer is 1. This is not an identity but a trick of giving the coefficient the form of $x$ ; it appears in the same kind of questions, so learn it here.

Coefficient	Effect on a pair of terms
$c = x + 1$	$x^{n} - c x^{n - 1} = - x^{n - 1}$
$c = x - 1$	$x^{n} - c x^{n - 1} = x^{n - 1}$

Here: $c$ = the repeating coefficient, $x$ = the given value, $n$ = the power. For $c = x + 1$ the next term comes with the opposite sign and the pair cancels; for $c = x - 1$ all later terms are negative and the power drops by one each time. A term that falls outside the pattern is set aside and added at the end.

Spot itSpot it: a value of

x

is given (often between 7 and 20), the powers go up to 4, 5 or 6, and several terms share one coefficient (equal to

x + 1

x - 1

Steps

Find the repeating coefficient and compare it with $x \pm 1$ .
Write $(x + 1)$ or $(x - 1)$ in place of each such coefficient.
Work from the left, cancelling two terms at a time.
Add at the end the term that did not fit the pattern.

Example 30 · Basic

For

x = 9

, find

x^{4} - 10 x^{3} + 10 x^{2} - 10 x + 10

Solution, step by step

1 $10 = 9 + 1 = x + 1$ , so write $(x + 1)$ in place of every 10.
2 $x^{4} - (x + 1) x^{3} = - x^{3}$ ; $- x^{3} + (x + 1) x^{2} = x^{2}$ ; $x^{2} - (x + 1) x = - x$ ; $- x + (x + 1) = 1$ .
3Answer 1. Check: $6561 - 7290 + 810 - 90 + 10 = 1$ ✓.

Answer 1

Example 31 · Standard

For

x = 19

, find

x^{5} - 20 x^{4} + 20 x^{3} - 20 x^{2} + 20 x - 3

Solution, step by step

1 $20 = x + 1$ . The five-power chain: $x^{5} - (x + 1) x^{4} = - x^{4}$ ; $- x^{4} + (x + 1) x^{3} = x^{3}$ ; $x^{3} - (x + 1) x^{2} = - x^{2}$ ; $- x^{2} + (x + 1) x = x$ .
2The chain leaves $x = 19$ ; the $- 3$ was separate.
3 $19 - 3 = 16$ .

Answer 16

If the chain's top power is odd the result is

+ x

, if even

- x

(when

c = x + 1

Example 32 · Standard

For

x = 8

, find

x^{4} - 7 x^{3} - 7 x^{2} - 7 x + 11

Solution, step by step

1 $7 = x - 1$ . Now all later terms are negative: $x^{4} - (x - 1) x^{3} = x^{3}$ .
2 $x^{3} - (x - 1) x^{2} = x^{2}$ , then $x^{2} - (x - 1) x = x$ .
3The chain $= x = 8$ ; $8 + 11 = 19$ . Check: $4096 - 3584 - 448 - 56 + 11 = 19$ ✓.

Answer 19

Example 33 · Advanced

For

x = 11

, find

x^{4} - 12 x^{3} + 14 x^{2} - 12 x + 9

Solution, step by step

1 $12 = x + 1$ , but the middle coefficient is 14, which is $12 + 2$ . Write it as $12 x^{2} + 2 x^{2}$ .
2 $x^{4} - (x + 1) x^{3} + (x + 1) x^{2} - (x + 1) x = - x$ (a chain of four terms).
3What remains: $- x + 2 x^{2} + 9 = - 11 + 242 + 9 = 240$ .

Answer 240

Split the term that does not fit into two parts.

Practice

BasicTry it yourself first

For

x = 6

, find the value of

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

BasicTry it yourself first

For

x = 7

, find the value of

x^{4} - 8 x^{3} + 8 x^{2} - 8 x

StandardTry it yourself first

For

x = 18

, find the value of

x^{5} - 19 x^{4} + 19 x^{3} - 19 x^{2} + 19 x + 6

StandardTry it yourself first

For

x = 13

, find the value of

x^{4} - 12 x^{3} - 12 x^{2} - 12 x

StandardTry it yourself first

For

x = 8

, find the value of

x^{6} - 9 x^{5} + 9 x^{4} - 9 x^{3} + 9 x^{2} - 9 x - 5

StandardTry it yourself first

For

x = 10

, find the value of

x^{5} - 9 x^{4} - 9 x^{3} - 9 x^{2} - 9 x + 2

AdvancedTry it yourself first

For

x = 15

, find

x^{4} - 16 x^{3} + 19 x^{2} - 16 x + 7

AdvancedTry it yourself first

For

x = 9

, find

x^{4} - 10 x^{3} + 8 x^{2} - 10 x + 4

AdvancedTry it yourself first

For

x = 17

, find the value of

x^{6} - 18 x^{5} + 18 x^{4} - 18 x^{3} + 18 x^{2} - 18 x + 12

AdvancedTry it yourself first

For

x = 16

, find the value of

x^{5} - 15 x^{4} - 15 x^{3} - 15 x^{2} - 15 x - 9

Part 2: three letters

Identities in

a, b, c

$(a + b + c)^{2}$ , $a^{3} + b^{3} + c^{3} - 3 ab c$ and the special results when $a + b + c = 0$ .

The most exam questions come from this part. The formulas look long, but all three share one frame: the sum $a + b + c$ , the pair-sum $ab + b c + c a$ and the sum of squares $a^{2} + b^{2} + c^{2}$ .

$(a + b + c)^{2}$ and what it is used for

The square of a three-letter sum

The nine pieces of $(a + b + c)^{2}$ , expanding, and getting the third of sum, pair-sum and sum of squares from the other two.

The idea

$(a + b + c)^{2}$ is a square with side $a + b + c$ . Cut it along both directions at $a, b, c$ and you get $3 \times 3 = 9$ pieces: three squares $a^{2}, b^{2}, c^{2}$ (on the diagonal) and six rectangles — $ab$ twice, $b c$ twice, $c a$ twice. So $(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab + 2 b c + 2 c a$ .

◭ Diagram

Three squares on the diagonal (blue, green, saffron); the other six rectangles come in pairs, one colour per pair

∑ Formula

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

Square of a three-letter sum

$a, b, c$: three numbers or expressions (with their signs)
$ab + b c + c a$: the sum of the products of the three pairs

It has three quantities: the sum $S = a + b + c$ , the pair-sum $q = ab + b c + c a$ and the sum of squares $T = a^{2} + b^{2} + c^{2}$ . The formula says $S^{2} = T + 2 q$ . Knowing any two gives the third: $q = \frac{S ^{2} - T}{2}$ , $T = S^{2} - 2 q$ , $S = \pm T + 2 q$ . A minus sign on a term gives minus in the pairs it is in: $(a + b - c)^{2} = a^{2} + b^{2} + c^{2} + 2 ab - 2 b c - 2 c a$ .

Spot itSpot it: a sum of three letters (or a squared bracket) with

ab + b c + c a

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

mentioned. When the sum is zero,

q = - \frac{T}{2}

Steps

Write which of $S, q, T$ are given.
Put the two known values in $S^{2} = T + 2 q$ .
Isolate the third (mind the factor 2).
When finding $S$ , check the sign: if the letters are positive then $S > 0$ .

Example 34 · Basic

Expand

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

Solution, step by step

1 $a = x$ , $b = 2 y$ , $c = - 3 z$ . Squares: $x^{2}, 4 y^{2}, 9 z^{2}$ (all positive, as $(- 3 z)^{2} = 9 z^{2}$ ).
2 $2 ab = 2 \times x \times 2 y = 4 x y$ ; $2 b c = 2 \times 2 y \times (- 3 z) = - 12 yz$ ; $2 c a = 2 \times (- 3 z) \times x = - 6 x z$ .
3 $x^{2} + 4 y^{2} + 9 z^{2} + 4 x y - 12 yz - 6 x z$ .

Answer $x^{2} + 4 x y - 6 x z + 4 y^{2} - 12 yz + 9 z^{2}$

Check

x = y = z = 1

(1 + 2 - 3)^{2} = 0

and

1 + 4 + 9 + 4 - 12 - 6 = 0

✓.

Example 35 · Standard

a + b + c = 11

and

a^{2} + b^{2} + c^{2} = 45

, find

ab + b c + c a

Solution, step by step

1 $S = 11$ , $T = 45$ . Formula $S^{2} = T + 2 q$ .
2 $121 = 45 + 2 q \Rightarrow 2 q = 76$ .
3 $q = 38$ . Check: $a = 5, b = 4, c = 2$ : sum 11, squares $25 + 16 + 4 = 45$ , pairs $20 + 8 + 10 = 38$ ✓.

Answer 38

Example 36 · Standard

a, b, c > 0

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

and

ab + b c + c a = 19

. Find

a + b + c

Solution, step by step

1 $(a + b + c)^{2} = 26 + 2 \times 19 = 64$ .
2Square root: $\pm 8$ . The sum of positive numbers is positive, so $a + b + c = 8$ .
3Check: $(4, 3, 1)$ — squares $16 + 9 + 1 = 26$ ✓, pairs $12 + 3 + 4 = 19$ ✓.

Answer 8

Example 37 · Advanced

a + b + c = 0

and

a^{2} + b^{2} + c^{2} = 38

, find

ab + b c + c a

and

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

Solution, step by step

1 $0 = 38 + 2 q \Rightarrow q = - 19$ .
2 $(a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 2 (a^{2} + b^{2} + c^{2}) - 2 (ab + b c + c a)$ (each $a^{2}$ appears twice, each $ab$ once with a minus, doubled).
3 $= 2 \times 38 - 2 \times (- 19) = 76 + 38 = 114$ . Check with $(5, - 2, - 3)$ : $49 + 1 + 64 = 114$ ✓.

Answer $- 19$ ; $114$

The second result will return in the next method as

E = \frac{1}{2} \sum (a - b)^{2}

Practice

BasicTry it yourself first

Expand

(x + 2 y + 3 z)^{2}

BasicTry it yourself first

Expand

(2 x - y + z)^{2}

StandardTry it yourself first

Expand

(3 x - 2 y - z)^{2}

BasicTry it yourself first

a + b + c = 11

and

a^{2} + b^{2} + c^{2} = 45

, find

ab + b c + c a

StandardTry it yourself first

a + b + c = 11

and

a^{2} + b^{2} + c^{2} = 59

, find

ab + b c + c a

StandardTry it yourself first

a + b + c = 9

and

ab + b c + c a = 8

, find

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

StandardTry it yourself first

For

a, b, c > 0

a^{2} + b^{2} + c^{2} = 42

and

ab + b c + c a = 29

, find

a + b + c

StandardTry it yourself first

For

a, b, c > 0

a^{2} + b^{2} + c^{2} = 49

and

ab + b c + c a = 36

, find

a + b + c

StandardTry it yourself first

a + b + c = 0

and

a^{2} + b^{2} + c^{2} = 74

, find

ab + b c + c a

AdvancedTry it yourself first

a + b + c = 0

and

a^{2} + b^{2} + c^{2} = 134

, find

ab + b c + c a

BasicTry it yourself first

For

x = 4, y = 3, z = 2

, find

x^{2} + y^{2} + z^{2} + 2 x y + 2 yz + 2 z x

AdvancedTry it yourself first

a + b + c = 10

and

ab + b c + c a = - 11

, find

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

AdvancedTry it yourself first

a + b + c = 13

and

a^{2} + b^{2} + c^{2} = 221

, find

ab + b c + c a

AdvancedTry it yourself first

a + b + c = 0

and

a^{2} + b^{2} + c^{2} = 218

, find

ab + b c + c a

The $a^{3} + b^{3} + c^{3} - 3 ab c$ identity

and

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

The three-letter formula, its half-sum-of-squared-gaps form, and questions with numbers in AP, with sums and with ratios.

The idea

First see $a^{3} + b^{3} + c^{3} - 3 ab c$ as $(a + b + c) \times E$ , where $E = a^{2} + b^{2} + c^{2} - ab - b c - c a$ . Check by multiplying: in $(a + b + c) E$ we get $a \cdot a^{2} = a^{3}$ ; the mixed terms like $- a^{2} b$ and $+ a^{2} b$ come in cancelling pairs; what survives is $a^{3} + b^{3} + c^{3}$ and $- 3 ab c$ . Second, $E$ can be written as $\frac{1}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}]$ , because on expanding the three squares each $a^{2}$ appears twice and each $- 2 ab$ once. Squares are never negative, so $E \geq 0$ , and $E = 0$ only when $a = b = c$ .

◭ Diagram

The pink strips are the gaps between bars; when the three bars become equal the gaps vanish

∑ Formula

a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a) = \frac{1}{2} (a + b + c) [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}]

The three-letter cube formula

$a + b + c$: the sum of the three
$E$: $a^{2} + b^{2} + c^{2} - ab - b c - c a$ : half the sum of the squares of the three gaps
$(a - b), (b - c), (c - a)$: the gaps of the three pairs

The benefit: when the three numbers are close together (like 498, 495, 493), take squares of the small gaps instead of big cubes. And when three numbers are in AP (common difference $d$ ) the gaps are $d, d, 2 d$ , so $E = \frac{1}{2} (d^{2} + d^{2} + 4 d^{2}) = 3 d^{2}$ .

Spot itSpot it: a direct mention of

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

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

; three close numbers; or a ratio like

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

Steps

If the numbers are given: find the gaps $a - b, b - c, c - a$ and compute $E$ .
If $S, T$ are given: $q = \frac{S ^{2} - T}{2}$ , then $E = T - q$ .
The value is $S \times E$ .
If you need $ab c$ , put the given sum of cubes into $a^{3} + b^{3} + c^{3} - 3 ab c$ and solve.

Example 38 · Basic

For

a = 64, b = 61, c = 59

, find

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

Solution, step by step

1Gaps: $a - b = 3$ , $b - c = 2$ , $c - a = - 5$ .
2Their squares: $9, 4, 25$ ; total $38$ .
3 $E = \frac{1}{2} \times 38 = 19$ . (Squaring 64, 61, 59 directly would take much longer.)

Answer 19

Example 39 · Standard

For

a = 12, b = 10, c = 8

, find

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

Solution, step by step

1 $a + b + c = 30$ .
2 $E = \frac{1}{2} (2^{2} + 2^{2} + 4^{2}) = \frac{1}{2} \times 24 = 12$ (the numbers are in AP with $d = 2$ , $E = 3 d^{2} = 12$ ✓).
3Value $= 30 \times 12 = 360$ . Check: $1728 + 1000 + 512 - 3 \times 960 = 3240 - 2880 = 360$ ✓.

Answer 360

Example 40 · Standard

a + b + c = 12

and

a^{2} + b^{2} + c^{2} = 50

, find

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

Solution, step by step

1 $ab + b c + c a = \frac{1 2 ^{2} - 50}{2} = \frac{94}{2} = 47$ .
2 $E = 50 - 47 = 3$ .
3Value $= 12 \times 3 = 36$ . Check: $(5, 4, 3)$ : $125 + 64 + 27 - 180 = 36$ ✓.

Answer 36

Example 41 · Advanced

a + b + c = 9

a^{2} + b^{2} + c^{2} = 29

and

a^{3} + b^{3} + c^{3} = 99

, find

ab c

Solution, step by step

1 $q = \frac{81 - 29}{2} = 26$ , so $E = 29 - 26 = 3$ .
2 $a^{3} + b^{3} + c^{3} - 3 ab c = 9 \times 3 = 27$ .
3 $99 - 3 ab c = 27 \Rightarrow 3 ab c = 72 \Rightarrow ab c = 24$ . Check: $(4, 3, 2)$ ✓.

Answer 24

Example 42 · Advanced

For

a = 31, b = 29, c = 28

, find

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

Solution, step by step

1The second form: numerator $= \frac{1}{2} (a + b + c) [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}]$ .
2The denominator is that same bracket, so it cancels.
3The answer is $\frac{1}{2} (31 + 29 + 28) = \frac{1}{2} \times 88 = 44$ .

Answer 44

Remember this relation: the ratio is always

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

Practice

BasicTry it yourself first

For

a = 86, b = 83, c = 80

, find

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

BasicTry it yourself first

For

a = 305, b = 302, c = 301

, find

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

StandardTry it yourself first

For

a = 25, b = 23, c = 21

, find

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

StandardTry it yourself first

For

a = 104, b = 101, c = 100

, find

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

StandardTry it yourself first

a + b + c = 12

and

a^{2} + b^{2} + c^{2} = 50

, find

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

StandardTry it yourself first

a + b + c = 15

and

a^{2} + b^{2} + c^{2} = 93

, find

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

AdvancedTry it yourself first

a + b + c = 11

a^{2} + b^{2} + c^{2} = 49

and

a^{3} + b^{3} + c^{3} = 251

, find

ab c

AdvancedTry it yourself first

a + b + c = 13

a^{2} + b^{2} + c^{2} = 75

and

a^{3} + b^{3} + c^{3} = 469

, find

ab c

StandardTry it yourself first

Three numbers are in AP with middle number 20 and common difference 3. Find the sum of their cubes minus three times their product.

AdvancedTry it yourself first

Three numbers are in AP with middle number 45 and common difference 4. Find the sum of their cubes minus three times their product.

StandardTry it yourself first

For

a = 40, b = 38, c = 37

, find

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

AdvancedTry it yourself first

For

a = 56, b = 55, c = 51

, find

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

AdvancedTry it yourself first

For

a = 702, b = 697, c = 696

, find

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

AdvancedTry it yourself first

For

a = 305, b = 303, c = 301

, find

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

When the sum is zero: the special cases

a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3 ab c

The zero-sum move, the forms with a minus, cubes of differences, cube-root conditions and $E = 0 \Rightarrow a = b = c$ .

The idea

The last identity says $a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) \cdot E$ . The right side can be zero in two ways: either the first factor $a + b + c = 0$ , or the second, $E = 0$ , which means $a = b = c$ . All the special results come from this.

◭ Diagram

Walk on the number line from 0:

+ 5

, then

- 2

, then

- 3

— you are back at 0

Condition	Result
$a + b + c = 0$	$a^{3} + b^{3} + c^{3} = 3 ab c$
$a + b - c = 0$	$a^{3} + b^{3} - c^{3} = - 3 ab c$
$a = b = c$ (or $E = 0$ )	$a^{3} + b^{3} + c^{3} - 3 ab c = 0$
$a^{3} + b^{3} + c^{3} = 3 ab c$ , and $a, b, c > 0$	$a = b = c$ (because $a + b + c \neq = 0$ )
$a^{2} + b^{2} = c^{2}$ (letters $a^{2}, b^{2}, - c^{2}$ )	$a^{6} + b^{6} - c^{6} = - 3 a^{2} b^{2} c^{2}$
$(x - y) + (y - z) + (z - x) = 0$	$(x - y)^{3} + (y - z)^{3} + (z - x)^{3} = 3 (x - y) (y - z) (z - x)$
$3 x + 3 y + 3 z = 0$	$x + y + z = 3 3 x yz$

Here: $E = a^{2} + b^{2} + c^{2} - ab - b c - c a$ . In the second row the first row is used with $c$ replaced by $- c$ : $a + b + (- c) = 0$ . In the differences row the three “letters” are $x - y, y - z, z - x$ , and they always add to zero. Trap: when $a^{3} + b^{3} + c^{3} = 3 ab c$ and $a, b, c$ are not stated to be positive, there are exactly two possibilities — $a + b + c = 0$ or $a = b = c$ .

Spot itSpot it: “

a + b + c = 0

”; “

a + b = c

”; three brackets whose sum has each letter once with

+

and once with

-

; a sum of cube roots; or “

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

”.

Steps

Treat the three terms as letters (or new letters $A, B, C$ ) and test their sum.
If the sum is zero, the sum of cubes $= 3 \times$ the product.
Carry the sign of any subtracted term along.
Finally put the result into what the question asks (a fraction, a ratio) and cancel.

Example 43 · Basic

For

a = 3.7, b = 2.8, c = - 6.5

, find

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

Solution, step by step

1 $3.7 + 2.8 - 6.5 = 0$ , so the sum is zero.
2 $a^{3} + b^{3} + c^{3} = 3 ab c = 3 \times 3.7 \times 2.8 \times (- 6.5)$ .
3 $3.7 \times 2.8 = 10.36$ ; $10.36 \times 6.5 = 67.34$ ; $\times 3 = 202.02$ ; the sign is negative: $- 202.02$ .

Answer $- 202.02$

This is far quicker than cubing three decimals and adding.

Example 44 · Standard

For

p = 1.7, q = 2.9, r = 4.6

, find

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

Solution, step by step

1 $1.7 + 2.9 = 4.6$ , i.e. $p + q - r = 0$ , or $p + q + (- r) = 0$ .
2With letters $p, q, - r$ : $p^{3} + q^{3} + (- r)^{3} = 3 pq (- r)$ , so $p^{3} + q^{3} - r^{3} = - 3 pq r$ .
3 $1.7 \times 2.9 = 4.93$ ; $4.93 \times 4.6 = 22.678$ ; $\times 3 = 68.034$ . The answer is $- 68.034$ .

Answer $- 68.034$

Example 45 · Standard

For

x = 8, y = 5, z = 2

, find

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

Solution, step by step

1The three gaps: $3, 3, - 6$ ; their sum is $0$ .
2The sum of cubes $= 3 \times 3 \times 3 \times (- 6) = - 162$ .
3Directly: $27 + 27 - 216 = - 162$ ✓.

Answer −162

Example 46 · Advanced

a, b, c > 0

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

and

a + b + c = 15

. Find

ab c

Solution, step by step

1 $(a + b + c) \cdot E = 0$ and $a + b + c = 15 \neq = 0$ , so $E = 0$ .
2 $E = \frac{1}{2} \sum (a - b)^{2} = 0$ only when every gap is zero: $a = b = c$ .
3All three are $5$ , so $ab c = 125$ .

Answer 125

Notice that the positivity condition removed the “

a + b + c = 0

” option.

Example 47 · Advanced

For

x = 9, y = 4, z = 5

, find

(x - 6)^{3} + (y - 6)^{3} + (z - 6)^{3}

Solution, step by step

1 $x + y + z = 18 = 3 \times 6$ , so the deviations $x - 6, y - 6, z - 6$ add to $0$ .
2Deviations: $3, - 2, - 1$ .
3Sum of cubes $= 3 \times 3 \times (- 2) \times (- 1) = 18$ . Directly: $27 - 8 - 1 = 18$ ✓.

Answer 18

Practice

BasicTry it yourself first

For

a = 7, b = - 4, c = - 3

, find

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

BasicTry it yourself first

For

a = 12, b = - 5, c = - 7

, find

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

StandardTry it yourself first

For

p = 3, q = 8, r = 11

, find

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

StandardTry it yourself first

For

p = 3, q = 4, r = 7

, find

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

StandardTry it yourself first

For

x = 9, y = 6, z = 1

, find

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

StandardTry it yourself first

For

x = 14, y = 3, z = 10

, find

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

StandardTry it yourself first

For

a, b, c > 0

with

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

and

a + b + c = 18

, find

ab c

StandardTry it yourself first

For

a, b, c > 0

with

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

, find

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

AdvancedTry it yourself first

3 x + 3 y + 3 z = 0

and

x yz = - 27, 000

, find

x + y + z

AdvancedTry it yourself first

3 x + 3 y + 3 z = 0

and

x yz = - 5, 92, 704

, find

x + y + z

AdvancedTry it yourself first

For

x = 12, y = 5, z = 7

, find

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

AdvancedTry it yourself first

For

x = 17, y = 5, z = 8

, find

(x - 10)^{3} + (y - 10)^{3} + (z - 10)^{3}

AdvancedTry it yourself first

a + b + c = 0

and

a, b, c \neq = 0

, what is the value of

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

AdvancedTry it yourself first

For

a, b, c > 0

with

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

and

a + b + c = 27

, find

ab c

StandardTry it yourself first

a, b, c

are natural numbers with

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

. Which of these can be

a + b + c

AdvancedTry it yourself first

For

p = 5, q = 12, r = 13

(note

p^{2} + q^{2} = r^{2}

), find

p^{6} + q^{6} - r^{6}

Part 3: questions with conditions

Symmetric sums ·

a^{4} + a^{2} b^{2} + b^{4}

· sum of squares zero

Some values are given and another expression is asked: here identities act as bridges.

Exams ask these the most: “this is given, find that”. You never need the letters separately; choosing the right identity is the whole job.

Symmetric sums of three letters

p = a + b + c, q = ab + b c + c a, r = ab c

A recipe, from the three basic quantities $p, q, r$ , for $a^{2} + b^{2} + c^{2}$ , $a^{3} + b^{3} + c^{3}$ , $(a + b) (b + c) (c + a)$ , $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ and more.

The idea

Expressions in three letters that do not change when the letters are swapped (symmetric ones) can always be built from three basic quantities: the sum $p = a + b + c$ , the pair-sum $q = ab + b c + c a$ and the product $r = ab c$ . (These are exactly what you get from the three roots $a, b, c$ of $t^{3} - p t^{2} + qt - r = 0$ .) So however long a question is, we ask: “how do I write this with $p, q, r$ ?”. Our earlier methods already gave the recipes:

Expression	In $p, q, r$	Where from
$a^{2} + b^{2} + c^{2}$	$p^{2} - 2 q$	expanding $(a + b + c)^{2}$
$(a - b)^{2} + (b - c)^{2} + (c - a)^{2}$	$2 p^{2} - 6 q$	expanding the squares
$a^{3} + b^{3} + c^{3}$	$p^{3} - 3 pq + 3 r$	$a^{3} + b^{3} + c^{3} - 3 ab c = p (p^{2} - 3 q)$
$(a + b) (b + c) (c + a)$	$pq - r$	expanding $(p - a) (p - b) (p - c)$
$\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$	$\frac{q}{r}$	common denominator $ab c$
$a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}$	$q^{2} - 2 p r$	expanding $(ab + b c + c a)^{2}$
$a^{4} + b^{4} + c^{4}$	$(p^{2} - 2 q)^{2} - 2 (q^{2} - 2 p r)$	square of the sum of squares

Here: $p = a + b + c$ (sum), $q = ab + b c + c a$ (pair-sum), $r = ab c$ (product). The recipe for $a^{3} + b^{3} + c^{3}$ comes from the three-letter identity: $a^{3} + b^{3} + c^{3} = 3 ab c + p (a^{2} + b^{2} + c^{2} - q) = 3 r + p (p^{2} - 3 q)$ . In $(a + b) (b + c) (c + a)$ each bracket is “ $p$ minus one letter”; so the product is $(p - a) (p - b) (p - c) = p^{3} - p^{3} + pq - r$ . One more link: $(a + b + c)^{3} = a^{3} + b^{3} + c^{3} + 3 (a + b) (b + c) (c + a)$ .

Spot itSpot it: three letters, with two or three of these given: sum, pair-sum, product, sum of squares, sum of cubes; and another symmetric expression asked.

Steps

Translate the given quantities into $p, q, r$ .
Pick the recipe of the asked expression from the table (if missing, go back one step to a simpler expression).
Put in the values of $p, q, r$ and compute.
To check, try the formula on a small triple (like $1, 2, 4$ ).

Example 48 · Basic

For

a = 1, b = 2, c = 4

find

p, q, r

and check

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

with

p^{3} - 3 pq + 3 r

Solution, step by step

1 $p = 1 + 2 + 4 = 7$ ; $q = 2 + 8 + 4 = 14$ ; $r = 8$ .
2 $p^{3} - 3 pq + 3 r = 343 - 294 + 24 = 73$ .
3Directly: $1 + 8 + 64 = 73$ ✓. The formula fits.

Answer 73

Example 49 · Standard

a + b + c = 7

ab + b c + c a = 14

ab c = 8

, find

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

and

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

Solution, step by step

1 $(a + b) (b + c) (c + a) = pq - r = 7 \times 14 - 8 = 90$ .
2 $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{q}{r} = \frac{14}{8} = \frac{7}{4}$ .
3Check with $(1, 2, 4)$ : $3 \times 6 \times 5 = 90$ ✓ and $1 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4}$ ✓.

Answer $90$ ; $\frac{7}{4}$

Example 50 · Standard

a + b + c = 10

and

ab + b c + c a = 31

, find

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

Solution, step by step

1 $a^{2} + b^{2} + c^{2} = p^{2} - 2 q = 100 - 62 = 38$ .
2Total $= 2 \times 38 - 2 \times 31 = 76 - 62 = 14$ .
3Check $(2, 3, 5)$ : $1 + 4 + 9 = 14$ ✓.

Answer 14

Example 51 · Advanced

a + b + c = 7

ab + b c + c a = 14

ab c = 8

, find

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

and

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

Solution, step by step

1 $\sum a^{2} b^{2} = q^{2} - 2 p r = 196 - 2 \times 7 \times 8 = 196 - 112 = 84$ .
2 $\sum a^{2} = p^{2} - 2 q = 49 - 28 = 21$ .
3 $\sum a^{4} = 2 1^{2} - 2 \times 84 = 441 - 168 = 273$ . Check with $(1, 2, 4)$ : $1 + 16 + 256 = 273$ ✓.

Answer $84$ ; $273$

Example 52 · Advanced

a + b + c = 9

ab + b c + c a = 26

and

a^{3} + b^{3} + c^{3} = 99

, find

ab c

and

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

Solution, step by step

1 $99 = p^{3} - 3 pq + 3 r = 729 - 702 + 3 r = 27 + 3 r$ , so $3 r = 72$ and $r = 24$ .
2 $(a + b) (b + c) (c + a) = pq - r = 9 \times 26 - 24 = 210$ .
3Check with $(4, 3, 2)$ : $7 \times 5 \times 6 = 210$ ✓.

Answer $24$ ; $210$

Practice

BasicTry it yourself first

a + b + c = 9

ab + b c + c a = 23

ab c = 15

, find

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

StandardTry it yourself first

a + b + c = 11

ab + b c + c a = 36

ab c = 36

, find

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

StandardTry it yourself first

a + b + c = 11

ab + b c + c a = 34

ab c = 24

, find

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

StandardTry it yourself first

a + b + c = 14

ab + b c + c a = 59

ab c = 70

, find

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

StandardTry it yourself first

ab + b c + c a = 36

and

ab c = 36

, find

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

StandardTry it yourself first

ab + b c + c a = 29

and

ab c = 20

, find

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

StandardTry it yourself first

a + b + c = 17

and

ab + b c + c a = 87

, find

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

AdvancedTry it yourself first

a + b + c = 9

ab + b c + c a = 26

ab c = 24

, find

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

AdvancedTry it yourself first

a + b + c = 8

ab + b c + c a = 19

ab c = 12

, find

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

AdvancedTry it yourself first

a + b + c = 11

ab + b c + c a = 38

and

a^{3} + b^{3} + c^{3} = 197

, find

ab c

AdvancedTry it yourself first

a + b + c = 15

ab + b c + c a = 68

ab c = 96

, find

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

AdvancedTry it yourself first

a + b + c = 10

ab + b c + c a = 23

ab c = 14

, find

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

$a^{4} + a^{2} b^{2} + b^{4}$ and the squaring move

Fourth-power expressions through squares

Factorising $a^{4} + a^{2} b^{2} + b^{4}$ , $a^{4} + 4 b^{4}$ , and squaring a condition to reach fourth powers.

The idea

Fourth-power expressions are hard to handle head-on, but they are squares of squares. Two moves help. Move 1: in $a^{4} + a^{2} b^{2} + b^{4}$ add and subtract $a^{2} b^{2}$ to get $(a^{2} + b^{2})^{2} - a^{2} b^{2}$ . This is a difference of squares, so $= (a^{2} + b^{2} + ab) (a^{2} + b^{2} - ab)$ . Move 2: square a condition (like $x + y = s$ ), then square the result once more; each time a new term like $2 x y$ appears, which we treat as the unknown and solve for.

∑ Formula

a^{4} + a^{2} b^{2} + b^{4} = (a^{2} + ab + b^{2}) (a^{2} - ab + b^{2}), a^{4} + 4 b^{4} = (a^{2} + 2 ab + 2 b^{2}) (a^{2} - 2 ab + 2 b^{2})

Two special factorisations

$a, b$: any two numbers or expressions
$a^{2} \pm ab + b^{2}$: neither factor breaks further

∑ Formula

x^{4} + y^{4} = (x^{2} + y^{2})^{2} - 2 x^{2} y^{2}, x^{2} + y^{2} = (x + y)^{2} - 2 x y

The ladder up to fourth powers

$x + y$: the given sum
$x y$: the unknown product we solve for

A caution: when the unknown $x y$ comes out with two values from a quadratic, both need not fit real $x, y$ . For real $x, y$ we must have $(x - y)^{2} = (x + y)^{2} - 4 x y \geq 0$ ; this condition removes one value.

Spot itSpot it: the shape

a^{4} + a^{2} b^{2} + b^{4}

; or

x^{4} + y^{4}

with

x + y

given; or a question giving

a^{2} + ab + b^{2}

and also

a^{4} + \dots

Steps

Decide whether the expression looks like $a^{4} + a^{2} b^{2} + b^{4}$ or like $x^{4} + y^{4}$ .
For the first, write the factorisation and get the other factor by dividing by the given one.
For the second, write $x^{2} + y^{2}$ and then $x^{4} + y^{4}$ in terms of $x y$ .
Solve the resulting equation and check the reality condition.

Example 53 · Basic

a^{2} + b^{2} = 10

and

ab = 3

, find

a^{4} + a^{2} b^{2} + b^{4}

Solution, step by step

1 $a^{4} + a^{2} b^{2} + b^{4} = (a^{2} + b^{2})^{2} - a^{2} b^{2}$ .
2 $= 100 - 9 = 91$ .
3Check: $(a, b) = (3, 1)$ : $81 + 9 + 1 = 91$ ✓.

Answer 91

Example 54 · Standard

a^{2} + ab + b^{2} = 13

and

a^{4} + a^{2} b^{2} + b^{4} = 91

, find

ab

and

a^{2} + b^{2}

Solution, step by step

1 $91 = 13 \times (a^{2} - ab + b^{2})$ , so $a^{2} - ab + b^{2} = 7$ .
2Two equations: $a^{2} + b^{2} + ab = 13$ and $a^{2} + b^{2} - ab = 7$ .
3Adding: $a^{2} + b^{2} = 10$ ; subtracting: $2 ab = 6$ , i.e. $ab = 3$ .

Answer $ab = 3$ ; $a^{2} + b^{2} = 10$

Example 55 · Standard

Factorise

x^{4} + 4

into quadratics and use them to check

5^{4} + 4

Solution, step by step

1 $x^{4} + 4 = x^{4} + 4 x^{2} + 4 - 4 x^{2} = (x^{2} + 2)^{2} - (2 x)^{2}$ .
2A difference of squares: $(x^{2} + 2 x + 2) (x^{2} - 2 x + 2)$ .
3 $x = 5$ : $37 \times 17 = 629$ and $5^{4} + 4 = 625 + 4 = 629$ ✓.

Answer $(x^{2} + 2 x + 2) (x^{2} - 2 x + 2)$

With this move a number like

x^{4} + 4

is never prime (for

x > 1

Example 56 · Advanced

For real

x, y

with

x + y = 4

and

x^{4} + y^{4} = 82

, find

x y

Solution, step by step

1 $p = x y$ . $x^{2} + y^{2} = 16 - 2 p$ ; $x^{4} + y^{4} = (16 - 2 p)^{2} - 2 p^{2} = 256 - 64 p + 2 p^{2}$ .
2 $256 - 64 p + 2 p^{2} = 82 \Rightarrow p^{2} - 32 p + 87 = 0 \Rightarrow (p - 3) (p - 29) = 0$ .
3Reality: $(x - y)^{2} = 16 - 4 p \geq 0 \Rightarrow p \leq 4$ . So $p = 29$ is out; $x y = 3$ . (Check: $x = 3, y = 1$ ✓)

Answer 3

Learn to discard the root that does not fit real numbers.

Example 57 · Advanced

For

a = 3, b = 1

, simplify

\frac{a ^{4} + b ^{4}}{a ^{2} + b ^{2} - 2 ab}

Solution, step by step

1 $a^{4} + b^{4} = (a^{2} + b^{2})^{2} - 2 a^{2} b^{2} = (a^{2} + b^{2})^{2} - (2 ab)^{2}$ .
2A difference of squares: $(a^{2} + b^{2} + 2 ab) (a^{2} + b^{2} - 2 ab)$ .
3The denominator cancelled: left with $a^{2} + b^{2} + 2 ab = 10 + 32$ . Check: $\frac{82}{10 - 3 2} = \frac{82 ( 10 + 3 2 )}{100 - 18} = 10 + 32$ ✓.

Answer $10 + 32$

Practice

BasicTry it yourself first

a^{2} + b^{2} = 5

and

ab = 2

, find

a^{4} + a^{2} b^{2} + b^{4}

BasicTry it yourself first

a^{2} + b^{2} = 25

and

ab = 12

, find

a^{4} + a^{2} b^{2} + b^{4}

StandardTry it yourself first

a^{2} + ab + b^{2} = 21

and

a^{4} + a^{2} b^{2} + b^{4} = 273

, find

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

StandardTry it yourself first

a^{2} + ab + b^{2} = 39

and

a^{4} + a^{2} b^{2} + b^{4} = 741

, find

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

StandardTry it yourself first

a^{2} + ab + b^{2} = 49

and

a^{4} + a^{2} b^{2} + b^{4} = 931

, find

ab

StandardTry it yourself first

a^{2} + ab + b^{2} = 37

and

a^{4} + a^{2} b^{2} + b^{4} = 481

, find

a^{2} + b^{2}

StandardTry it yourself first

Break

x^{4} + 4

into a product of two quadratics.

StandardTry it yourself first

Break

x^{4} + 64

into a product of two quadratics.

AdvancedTry it yourself first

For real

x, y

with

x + y = 3

and

x^{4} + y^{4} = 17

, find

x y

AdvancedTry it yourself first

For real

x, y

with

x + y = 6

and

x^{4} + y^{4} = 626

, find

x y

AdvancedTry it yourself first

a^{2} + ab + b^{2} = 93

and

a^{4} + a^{2} b^{2} + b^{4} = 3, 441

, find

ab

AdvancedTry it yourself first

a^{2} + b^{2} = 74

and

ab = 35

, find

a^{4} + a^{2} b^{2} + b^{4}

When a sum of squares is zero

Each square is zero on its own

Using “a square is never negative” to put each bracket equal to zero, and making squares to find the values of the letters.

The idea

The square of any real number is zero or positive, never negative. So if the sum of several squares is zero, no square can be positive — the others could not bring it back to zero. Hence every square is zero on its own. That is why $(x - 4)^{2} + (y + 6)^{2} = 0$ gives straight away $x = 4, y = - 6$ — two values from one equation.

◭ Diagram

All three bars shrink to zero height; none can go below zero

∑ Formula

A^{2} + B^{2} + C^{2} = 0 ⟹ A = 0, B = 0, C = 0

Sum of squares equal to zero

$A, B, C$: any real expressions (like $x - 4$ , $2 x + y - 5$ )

If the equation is not given as squares, make them: $x^{2} - 2 h x + h^{2} = (x - h)^{2}$ . Look at the group of terms for each letter and split the constant as needed. Example: in $x^{2} + y^{2} - 8 x + 6 y + 25 = 0$ we have $25 = 16 + 9$ , so $(x^{2} - 8 x + 16) + (y^{2} + 6 y + 9) = 0$ . When letters multiply each other (an $x y$ term) a square is of a linear form in two letters, like $(x + y)^{2}$ or $(2 x + y)^{2}$ . One more form of the rule: $x^{2} + y^{2} + z^{2} = x y + yz + z x \Rightarrow x = y = z$ .

Spot itSpot it: one equation, several letters, and the values of the letters (or a combination) are asked. The words “for real numbers” are also a hint. (Note: if the sum of squares is equal to a negative number there is no real solution.)

Steps

Bring all terms to one side so that the right side is $0$ .
Make a perfect square from the terms of each letter ( $x^{2} - 2 h x + h^{2}$ ); split the constant.
Put each bracket equal to zero separately.
Use the values to get the asked expression.

Example 58 · Basic

(x - 4)^{2} + (y + 6)^{2} + (z - 1)^{2} = 0

, find

x + y + z

and

x yz

Solution, step by step

1Each square is $\geq 0$ and their total is $0$ , so each is $0$ .
2 $x - 4 = 0$ , $y + 6 = 0$ , $z - 1 = 0$ , i.e. $x = 4, y = - 6, z = 1$ .
3 $x + y + z = 4 - 6 + 1 = - 1$ and $x yz = 4 \times (- 6) \times 1 = - 24$ .

Answer $- 1$ ; $- 24$

Example 59 · Standard

x^{2} + y^{2} + z^{2} - 2 x + 4 y - 6 z + 14 = 0

, find

x + y + z

Solution, step by step

1Split the constant $14 = 1 + 4 + 9$ and keep each letter's terms together: $(x^{2} - 2 x + 1) + (y^{2} + 4 y + 4) + (z^{2} - 6 z + 9) = 0$ .
2 $(x - 1)^{2} + (y + 2)^{2} + (z - 3)^{2} = 0$ , so $x = 1, y = - 2, z = 3$ .
3 $x + y + z = 1 - 2 + 3 = 2$ .

Answer 2

Example 60 · Standard

Solve

2 x^{2} + 2 x y + y^{2} + 6 x + 9 = 0

(

x, y

real).

Solution, step by step

1There is an $x y$ term, so make $(x + y)^{2} = x^{2} + 2 x y + y^{2}$ : what remains is $x^{2} + 6 x + 9$ , which is $(x + 3)^{2}$ .
2The equation: $(x + y)^{2} + (x + 3)^{2} = 0$ .
3 $x + 3 = 0 \Rightarrow x = - 3$ ; $x + y = 0 \Rightarrow y = 3$ . Check: $18 - 18 + 9 - 18 + 9 = 0$ ✓.

Answer $x = - 3, y = 3$

The shape of the square: the linear form that carries the

x y

term.

Example 61 · Advanced

(x + y - 3)^{2} + (y + z - 5)^{2} + (z + x - 6)^{2} = 0

, find

x + y + z

and

x yz

Solution, step by step

1All three zero: $x + y = 3$ , $y + z = 5$ , $z + x = 6$ .
2Adding gives $2 (x + y + z) = 14$ , so $x + y + z = 7$ .
3 $z = 7 - 3 = 4$ , $x = 7 - 5 = 2$ , $y = 7 - 6 = 1$ ; $x yz = 2 \times 1 \times 4 = 8$ .

Answer $7$ ; $8$

Example 62 · Advanced

For positive

x, y, z

with

x^{2} + y^{2} + z^{2} = x y + yz + z x

and

x^{2} + y^{2} + z^{2} = 147

, find

x + y + z

Solution, step by step

1 $x^{2} + y^{2} + z^{2} - x y - yz - z x = 0$ means $\frac{1}{2} [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}] = 0$ .
2Each square is zero, so $x = y = z$ .
3 $3 x^{2} = 147 \Rightarrow x^{2} = 49 \Rightarrow x = 7$ (positive); the sum is $21$ .

Answer 21

Practice

BasicTry it yourself first

For real

x, y, z

(x - 3)^{2} + (y + 2)^{2} + (z - 5)^{2} = 0

. Find the value of

x + y + z

BasicTry it yourself first

For real

x, y, z

(x + 1)^{2} + (y - 4)^{2} + (z - 2)^{2} = 0

. Find the value of

x yz

StandardTry it yourself first

For real

x, y

x^{2} + y^{2} - 8 x + 6 y + 25 = 0

. Find the value of

x^{2} + y^{2}

StandardTry it yourself first

For real

x, y, z

x^{2} + y^{2} + z^{2} - 4 x + 10 y - 6 z + 38 = 0

. Find

x + y + z

StandardTry it yourself first

For real

x, y, z

x^{2} + y^{2} + z^{2} + 6 x - 2 y - 12 z + 46 = 0

. Find

x yz

StandardTry it yourself first

For real

x, y, z

(x - 7)^{2} + (y - 2)^{2} + (z + 4)^{2} = 0

. Find the value of

x - y + z

StandardTry it yourself first

For real

x, y

2 x^{2} + 4 x y + 4 y^{2} - 4 x - 16 y + 20 = 0

. Find the value of

x + y

AdvancedTry it yourself first

For real

x, y

10 x^{2} - 12 x y + 4 y^{2} - 68 x + 40 y + 116 = 0

. Find the value of

x + y

StandardTry it yourself first

For real

x, y, z

(x + y - 7)^{2} + (y + z - 5)^{2} + (z + x - 8)^{2} = 0

. Find

x + y + z

AdvancedTry it yourself first

For real

x, y, z

(x + y - 8)^{2} + (y + z - 6)^{2} + (z + x - 10)^{2} = 0

. Find

x yz

AdvancedTry it yourself first

For positive

x, y, z

with

x^{2} + y^{2} + z^{2} = x y + yz + z x

and

x^{2} + y^{2} + z^{2} = 75

, find

x + y + z

AdvancedTry it yourself first

For positive

x, y, z

with

x^{2} + y^{2} + z^{2} = x y + yz + z x

and

x^{2} + y^{2} + z^{2} = 432

, find

x + y + z

Part 4: factors, remainders, least values and checking

Four more uses of identities

Reading right to left (factorising), the remainder theorem, least/greatest values by completing squares, and checking by substitution.

The last part is for questions where the identity does not show openly — but works behind the scenes all the same.

Factorising with identities

Reading the identities backwards

Which shape calls for which identity: common factor, squares, cubes, pairing and splitting the middle term.

The idea

Factorising is the same identity read in the opposite direction: from the expanded form to the product form. So every identity is a mould; we recognise the mould from the shape of the expression. The table below is the map.

Shape of the expression	Mould / identity	Factors
a common letter/number in all terms	take the common factor out	$k (\dots)$
two terms, both squares, with $-$	$a^{2} - b^{2}$	$(a + b) (a - b)$
three terms: square, $2 ab$ , square	$a^{2} \pm 2 ab + b^{2}$	$(a \pm b)^{2}$
$x^{2} + (m + n) x + mn$	sum $m + n$ , product $mn$	$(x + m) (x + n)$
two terms, both cubes	$a^{3} \pm b^{3}$	$(a \pm b) (a^{2} \mp ab + b^{2})$
$a^{4} + a^{2} b^{2} + b^{4}$	add and subtract a square	$(a^{2} + ab + b^{2}) (a^{2} - ab + b^{2})$
four terms	pair up for a common bracket	$(\dots) (\dots)$
three cubes and $- 3 ab c$	$a^{3} + b^{3} + c^{3} - 3 ab c$	$(a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$

Here: $a, b$ = any numbers or expressions; $m, n$ = the two numbers whose product is the last term and whose sum is the coefficient of $x$ . Order matters: first the common factor, then the identity that fits the shape, then the same rule again on each factor until nothing breaks further. Finally check by expanding (or by putting in one number).

Spot itSpot it: “factorise”, “is a factor of”, or a fraction whose numerator or denominator might cancel.

Steps

Find and take out a common factor.
Count the terms, look at the shape, and choose the mould from the table.
Put in $a$ and $b$ and write the factors.
Check each factor again to see whether it breaks further; then verify by expanding.

Example 63 · Basic

Factorise

18 x^{3} - 50 x

completely.

Solution, step by step

1Both terms share $2 x$ : $18 x^{3} - 50 x = 2 x (9 x^{2} - 25)$ .
2 $9 x^{2} - 25 = (3 x)^{2} - 5^{2}$ , a difference of squares.
3 $18 x^{3} - 50 x = 2 x (3 x + 5) (3 x - 5)$ .

Answer $2 x (3 x + 5) (3 x - 5)$

Check

x = 1

18 - 50 = - 32

and

2 \times 8 \times (- 2) = - 32

✓.

Example 64 · Standard

Factorise

x^{2} + 7 x + 12

and

x^{2} - 2 x - 15

Solution, step by step

1First: product $12$ , sum $7$ : of the pairs, $(3, 4)$ has $3 + 4 = 7$ ✓. So $(x + 3) (x + 4)$ .
2Second: product $- 15$ , sum $- 2$ : the pair $(3, - 5)$ has $3 + (- 5) = - 2$ ✓. So $(x + 3) (x - 5)$ .
3Check by expanding: $(x + 3) (x - 5) = x^{2} - 5 x + 3 x - 15 = x^{2} - 2 x - 15$ ✓.

Answer $(x + 3) (x + 4)$ ; $(x + 3) (x - 5)$

It is the same “sum and product” thinking as in the earlier part with two letters.

Example 65 · Standard

Factorise

a^{2} + 2 ab + b^{2} - 9

Solution, step by step

1The first three terms form a perfect square: $a^{2} + 2 ab + b^{2} = (a + b)^{2}$ .
2The expression $= (a + b)^{2} - 3^{2}$ , a difference of squares.
3 $= (a + b + 3) (a + b - 3)$ .

Answer $(a + b + 3) (a + b - 3)$

Example 66 · Advanced

Factorise

x^{3} - 3 x^{2} - 4 x + 12

completely.

Solution, step by step

1Four terms: pair them: $x^{2} (x - 3) - 4 (x - 3)$ .
2Take out the common bracket $(x - 3)$ : $(x - 3) (x^{2} - 4)$ .
3 $x^{2} - 4 = (x + 2) (x - 2)$ , so complete: $(x - 3) (x - 2) (x + 2)$ . Check: $x = 1$ : $1 - 3 - 4 + 12 = 6$ and $(- 2) (- 1) (3) = 6$ ✓.

Answer $(x - 3) (x - 2) (x + 2)$

Example 67 · Advanced

Factorise

x^{3} + 8 y^{3} + z^{3} - 6 x yz

Solution, step by step

1 $8 y^{3} = (2 y)^{3}$ and $6 x yz = 3 \times x \times 2 y \times z$ . So this is $A^{3} + B^{3} + C^{3} - 3 A BC$ with $A = x$ , $B = 2 y$ , $C = z$ .
2Factors: $(A + B + C) (A^{2} + B^{2} + C^{2} - A B - BC - C A)$ .
3 $A B = 2 x y$ , $BC = 2 yz$ , $C A = z x$ , so $(x + 2 y + z) (x^{2} + 4 y^{2} + z^{2} - 2 x y - 2 yz - z x)$ .

Answer $(x + 2 y + z) (x^{2} + 4 y^{2} + z^{2} - 2 x y - 2 yz - z x)$

Practice

BasicTry it yourself first

Factorise

x^{2} + 8 x + 15

StandardTry it yourself first

Factorise

x^{2} - 5 x - 36

StandardTry it yourself first

Factorise

12 x^{3} - 147 x

completely.

StandardTry it yourself first

Factorise

9 x^{2} + 12 x y + 4 y^{2} - 25

StandardTry it yourself first

Factorise

x^{3} - 4 x^{2} - 9 x + 36

AdvancedTry it yourself first

Factorise

x^{4} + 9 x^{2} + 81

StandardTry it yourself first

Which one is a factor of

27 x^{3} - 8

AdvancedTry it yourself first

Factorise

a^{3} + 27 b^{3} + 8 c^{3} - 18 ab c

AdvancedTry it yourself first

Factorise

x^{2} - 13 x + 42

AdvancedTry it yourself first

Factorise

x^{3} + 2 x^{2} - 25 x - 50

Remainder and factor theorem

The remainder without dividing

The remainder of $f (x) \div (x - r)$ is $f (r)$ ; finding unknown coefficients; remainders of powers; synthetic division.

The idea

When a polynomial $f (x)$ is divided by $(x - r)$ we get $f (x) = (x - r) Q (x) + R$ , where $R$ is a number (the remainder). This identity holds for every value of $x$ , so put $x = r$ : $(r - r) Q (r) = 0$ and what is left is $f (r) = R$ . So the remainder is just $f (r)$ . In particular, if $R = 0$ then $(x - r)$ is a factor. In a graph: where the curve crosses the $x$ -axis that value is a root of $f (x)$ and $(x - r)$ is a factor.

◭ Diagram

The curve is

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

; the red dot runs along the curve, the green circles are the roots

◭ Diagram

x = 4

the curve has height 3; that is the remainder, because

x^{2} - 6 x + 11 = (x - 4) (x - 2) + 3

Statement	Rule
remainder of $f (x) \div (x - r)$	$f (r)$
remainder of $f (x) \div (a x - b)$	$f (\frac{b}{a})$
$(x - r)$ is a factor	$f (r) = 0$
$x^{n} - a^{n}$	always divisible by $(x - a)$ ; by $(x + a)$ too if $n$ is even
$x^{n} + a^{n}$	divisible by $(x + a)$ if $n$ is odd
$f (x) \div (x^{2} + 1)$	put $x^{2} = - 1$
$f (x) \div (x^{4} + 1)$	put $x^{4} = - 1$

Here: $f (x)$ = the given polynomial, $r$ = the value that makes the divisor zero, $a, b$ = the numbers in $a x - b$ . The divisibility of $x^{n} \pm a^{n}$ is the algebraic form of the “ $a^{n} \pm b^{n}$ ” remark of the number-system chapter. Synthetic division (bring down a coefficient, multiply by $r$ , add to the next) is a short way to get the quotient; the last number is the remainder.

Spot itSpot it: “what is the remainder?”, “

(x - a)

is a factor, find

k

”, “factorise” (a cubic or higher polynomial), or a very big power under division.

Steps

Put the divisor equal to zero and find the value of $x$ (the root).
Put that value in the polynomial and compute: that is the remainder.
If there is an unknown $k$ , put the remainder equal to the given number (or $0$ for a factor) and solve for $k$ .
For a full factorisation test roots ( $\pm 1, \pm 2 \dots$ ), do synthetic division and break the leftover quadratic.

Example 68 · Basic

Find the remainder when

f (x) = 2 x^{3} - 5 x^{2} + x + 9

is divided by

(x - 2)

Solution, step by step

1 $x - 2 = 0 \Rightarrow x = 2$ .
2 $f (2) = 2 \times 8 - 5 \times 4 + 2 + 9 = 16 - 20 + 2 + 9$ .
3 $= 7$ , and that is the remainder.

Answer 7

Check (by division):

2 x^{3} - 5 x^{2} + x + 9 = (x - 2) (2 x^{2} - x - 1) + 7

✓.

Example 69 · Standard

(x + 2)

is a factor of

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

. Find

k

Solution, step by step

1It is a factor, so $f (- 2) = 0$ at $x = - 2$ .
2 $f (- 2) = - 8 + 4 k + 10 - 6 = 4 k - 4$ .
3 $4 k - 4 = 0 \Rightarrow k = 1$ . Check: $x^{3} + x^{2} - 5 x - 6 = (x + 2) (x^{2} - x - 3)$ ✓.

Answer 1

Example 70 · Standard

Find the remainder when

x^{37} + x^{12} + 9

is divided by

(x + 1)

Solution, step by step

1 $x + 1 = 0 \Rightarrow x = - 1$ .
2 $(- 1)^{37} = - 1$ (odd power) and $(- 1)^{12} = + 1$ (even power).
3Remainder $= - 1 + 1 + 9 = 9$ .

Answer 9

Example 71 · Advanced

(x - 1)

and

(x + 2)

are both factors of

x^{3} + a x^{2} + b x - 6

. Find

a

and

b

Solution, step by step

1 $f (1) = 0$ : $1 + a + b - 6 = 0 \Rightarrow a + b = 5$ .
2 $f (- 2) = 0$ : $- 8 + 4 a - 2 b - 6 = 0 \Rightarrow 2 a - b = 7$ .
3Adding gives $3 a = 12$ , so $a = 4$ and $b = 1$ . Check: $x^{3} + 4 x^{2} + x - 6 = (x - 1) (x + 2) (x + 3)$ ✓.

Answer $a = 4, b = 1$

Example 72 · Advanced

Factorise

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

completely.

Solution, step by step

1The divisors of the constant $- 6$ : $\pm 1, \pm 2, \pm 3, \pm 6$ . $f (1) = 1 - 6 + 11 - 6 = 0$ , so $(x - 1)$ is a factor.
2Synthetic division ( $r = 1$ ): $1, 1 - 6 = - 5, - 5 + 11 = 6, 6 - 6 = 0$ . The quotient is $x^{2} - 5 x + 6$ .
3 $x^{2} - 5 x + 6$ : sum $- 5$ , product $6$ — the numbers $- 2, - 3$ . Complete: $(x - 1) (x - 2) (x - 3)$ .

Answer $(x - 1) (x - 2) (x - 3)$

The three roots add to

6

, the negative of the coefficient of

x^{2}

✓.

Practice

BasicTry it yourself first

What is the remainder when

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

is divided by

x - 3

BasicTry it yourself first

What is the remainder when

2 x^{3} - 7 x + 4

is divided by

x + 2

StandardTry it yourself first

(x - 3)

is a factor of

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

. Find

k

StandardTry it yourself first

Find the remainder when

x^{41} + x^{20} + 6

is divided by

x + 1

StandardTry it yourself first

Find the remainder when

x^{30} + 5

is divided by

x^{2} + 1

StandardTry it yourself first

Find the remainder when

3 x^{2} - 2 x + 4

is divided by

(2 x - 1)

StandardTry it yourself first

What is the remainder when

x^{4} + 2 x^{3} - 5 x^{2} + 7 x - 3

is divided by

x - 2

AdvancedTry it yourself first

(x - 1)

and

(x - 2)

are both factors of

x^{3} + a x^{2} + b x + 8

. Find

a + b

StandardTry it yourself first

What is the quotient when

x^{3} - 10 x^{2} + 31 x - 30

is divided by

(x - 2)

AdvancedTry it yourself first

Factorise

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

completely.

AdvancedTry it yourself first

(x + 2)

is a factor of

x^{3} + k x^{2} + 2 x - 8

. Find

k

AdvancedTry it yourself first

Find the remainder when

x^{64} + x^{81} - 11

is divided by

x - 1

StandardTry it yourself first

Find the remainder when

x^{60} + 8

is divided by

x^{4} + 1

AdvancedTry it yourself first

Find the remainder when

x^{88} - 6

is divided by

x^{4} + 1

Least and greatest values

Completing the square ·

A + B \geq 2 A B

The least/greatest value of $a x^{2} + b x + c$ , squares in two letters, product with a fixed sum and sum with a fixed product.

The idea

Every square is $\geq 0$ and is zero only when its bracket is zero. So if an expression can be made “a square + a constant”, its least value is that constant (when the square is zero). This is called completing the square: $x^{2} - 6 x + 14 = (x - 3)^{2} + 5$ . Since $(x - 3)^{2} \geq 0$ the expression is $\geq 5$ , with equality at $x = 3$ . In a graph it is the lowest point of the curve (the vertex).

◭ Diagram

The red dot runs along the curve; the saffron dot is the lowest point

∑ Formula

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

Completing the square in general

$a, b, c$: coefficients of the quadratic ( $a \neq = 0$ )
$x = - \frac{b}{2 a}$: the value where the square is zero
$\frac{4 a c - b ^{2}}{4 a}$: the least ( $a > 0$ ) or greatest ( $a < 0$ ) value

Situation	Result
$a > 0$	least value $\frac{4 a c - b ^{2}}{4 a}$ (no greatest)
$a < 0$	greatest value $\frac{4 a c - b ^{2}}{4 a}$ (no least)
$a + b$ fixed, $a, b > 0$	$ab$ greatest $(\frac{a + b}{2})^{2}$ , when $a = b$
$ab$ fixed, $a, b > 0$	$a + b$ least $2 ab$ , when $a = b$
$(x - m) (x - n)$	least $- (\frac{m - n}{2})^{2}$ , at $x = \frac{m + n}{2}$

Here $A, B$ are positive numbers. Both statements come from one identity, $(A + B)^{2} - (A - B)^{2} = 4 A B$ (from part 1). If the sum is fixed, $4 A B = (A + B)^{2} - (A - B)^{2}$ is greatest when $(A - B)^{2} = 0$ . If instead $A B$ is fixed, $(A + B)^{2} = 4 A B + (A - B)^{2} \geq 4 A B$ , i.e. $A + B \geq 2 A B$ . Caution: the least value is really attained only if that equality is possible (if $A = B$ for some allowed value of the letter).

Spot itSpot it: “least/greatest value”, “smallest/largest”, an area to be greatest (perimeter fixed), or an expression like

x + \frac{1}{x}

Steps

Look at the sign of $a$ to decide whether a least or a greatest value exists.
Complete the square, or apply $\frac{4 a c - b ^{2}}{4 a}$ directly.
In a two-quantity question see whether the sum or the product is fixed.
Check that the equality condition (like $A = B$ ) is possible.

Example 73 · Basic

Find the least value of

y = x^{2} - 6 x + 14

and the

x

where it occurs.

Solution, step by step

1Complete the square: $x^{2} - 6 x + 9 = (x - 3)^{2}$ , so $y = (x - 3)^{2} + 5$ ( $14 = 9 + 5$ ).
2 $(x - 3)^{2} \geq 0$ , so $y \geq 5$ .
3Equality at $x = 3$ . Least value $5$ . (Formula: $\frac{4 \times 14 - 36}{4} = 5$ ✓.)

Answer $5$ ( $x = 3$ पर)

Example 74 · Standard

Find the greatest value of

y = - 2 x^{2} + 12 x - 7

Solution, step by step

1 $a = - 2 < 0$ , so a greatest value exists. $y = - 2 (x^{2} - 6 x) - 7 = - 2 [(x - 3)^{2} - 9] - 7$ .
2 $= - 2 (x - 3)^{2} + 18 - 7 = - 2 (x - 3)^{2} + 11$ .
3 $- 2 (x - 3)^{2} \leq 0$ , so $y \leq 11$ ; the greatest value is $11$ (at $x = 3$ ). Formula: $\frac{4 ( - 2 ) ( - 7 ) - 144}{4 ( - 2 )} = \frac{- 88}{- 8} = 11$ ✓.

Answer 11

Example 75 · Standard

Two positive numbers add to

26

. Find the greatest product and the numbers.

Solution, step by step

1 $4 ab = (a + b)^{2} - (a - b)^{2} = 676 - (a - b)^{2}$ .
2 $(a - b)^{2} \geq 0$ , so $4 ab \leq 676$ , i.e. $ab \leq 169$ .
3Equality when $a = b = 13$ . The greatest product is $169$ .

Answer $169$ ; $13, 13$

Example 76 · Advanced

For

x > 0

, find the least value of

3 x + \frac{12}{x}

Solution, step by step

1 $A = 3 x$ and $B = \frac{12}{x}$ are both positive and $A B = 36$ (the letter cancelled).
2 $A + B \geq 2 A B = 2 \times 6 = 12$ .
3Equality is possible at $3 x = \frac{12}{x} \Rightarrow x^{2} = 4 \Rightarrow x = 2$ . Least value $12$ . Check: $x = 2$ : $6 + 6 = 12$ ✓.

Answer 12

Example 77 · Advanced

Find the least value of

x^{2} + y^{2} - 8 x + 10 y + 45

Solution, step by step

1Make squares for both letters: $(x^{2} - 8 x + 16) + (y^{2} + 10 y + 25) + 4$ ( $45 = 16 + 25 + 4$ ).
2 $= (x - 4)^{2} + (y + 5)^{2} + 4$ . Both squares are $\geq 0$ .
3Both are zero at $x = 4, y = - 5$ ; the least value is $4$ .

Answer 4

Practice

BasicTry it yourself first

Find the least value of

y = x^{2} - 8 x + 19

BasicTry it yourself first

Find the least value of

y = 2 x^{2} - 12 x + 25

StandardTry it yourself first

Find the greatest value of

y = - x^{2} + 10 x - 9

StandardTry it yourself first

Find the value of

x

at which the least value of

y = 3 x^{2} + 12 x + 5

occurs.

StandardTry it yourself first

Find the least value of

x^{2} + y^{2} - 10 x + 6 y + 41

for real

x, y

StandardTry it yourself first

The sum of two positive numbers is

40

. Find the greatest value of their product.

StandardTry it yourself first

For

x > 0

, find the least value of

4 x + \frac{9}{x}

AdvancedTry it yourself first

For

x > 0

, find the least value of

5 x + \frac{20}{x}

StandardTry it yourself first

A wire 48 m long is bent into a rectangle. Find the greatest possible area in square metres.

AdvancedTry it yourself first

Find the least value of

(x - 13) (x - 3)

AdvancedTry it yourself first

Find the least value of

y = x^{2} - 7 x + 20

AdvancedTry it yourself first

Find the least value of

x^{2} + y^{2} + 12 x - 4 y + 37

for real

x, y

Checking by substitution

An identity is true for every value

Testing a formula or your own answer by putting in values, choosing among options and finding coefficients — and the traps in which even substitution deceives.

The idea

An identity means “true for every value”. That makes two jobs easy. (1) Testing: put in any pair of values; if the two sides differ, the formula or your answer is wrong — a single example is enough. (2) Choosing: among options, the right one gives the same value as the expression at every test value. The converse is not true: agreeing at one value does not prove it is correct.

Test values	Why good / bad
$a = 2, b = 1$	small, different, non-zero: most wrong formulas are caught here
$a = 3, b = 2$	a second test when two options tie on the first pair
$a = b$	poor: $a - b = 0$ wipes out many terms — a wrong formula may escape
$a = 0$ or $b = 0$	poor: one letter disappears
$a = 1, b = 1$	poor: powers of $1$ all look the same

Here: $a, b$ = the letters of the identity. Why several values may be needed: two polynomials of degree $n$ in $x$ that agree at $n + 1$ different values are the same identity — so in a question with unknown coefficients you need as many substitutions as there are unknowns. Caution: substitution is a way of checking, not a full proof; for exam options it is always enough.

Spot itSpot it: “which option is correct” where the options are expressions in letters; “which of these is not an identity”; “find

p, q

if both sides are equal for all

x

”; and when checking your own factorisation or expansion.

Steps

Choose test values in which no letter is zero, equal to the other, or $1$ .
Find the numerical value of the original expression and of every option.
If two options tie, try a second pair.
With unknown coefficients, put in easy values ( $x = 0, 1, - 1$ ) to form equations.

Example 78 · Basic

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

an identity? Test with

a = 2, b = 1

Solution, step by step

1Left side: $2^{2} + 1^{2} = 5$ .
2Right side: $(2 + 1)^{2} = 9$ .
3 $5 \neq = 9$ , so it is not an identity; (the gap $4 = 2 ab$ is exactly the missing term).

Answer Not an identity

Example 79 · Standard

What is

(a - 2 b)^{2} + 8 ab

equal to:

(a + 2 b)^{2}

(a - 2 b)^{2}

a^{2} + 4 b^{2}

a^{2} - 4 b^{2}

Solution, step by step

1Take $(a, b) = (1, 0)$ : the expression $= 1$ and the options give $1, 1, 1, 1$ — all equal, so this pair decides nothing.
2Take $(a, b) = (3, 1)$ : the expression $= (3 - 2)^{2} + 24 = 25$ . The options give: $25, 1, 13, 5$ .
3Only $(a + 2 b)^{2}$ gave 25. (By expanding: $a^{2} - 4 ab + 4 b^{2} + 8 ab = a^{2} + 4 ab + 4 b^{2}$ ✓)

Answer $(a + 2 b)^{2}$

The first pair had

b = 0

: a poor choice that made everything tie.

Example 80 · Standard

(x - 1) (x^{2} + p x + q) = x^{3} + x^{2} - 4 x + 2

holds for all

x

. Find

p

and

q

Solution, step by step

1 $x = 0$ : $(0 - 1) q = 2$ , so $q = - 2$ .
2 $x = 2$ : left $(2 - 1) (4 + 2 p + q) = 4 + 2 p - 2$ ; right $8 + 4 - 8 + 2 = 6$ .
3 $2 + 2 p = 6 \Rightarrow p = 2$ . Check $x = 1$ : left $0$ ; right $1 + 1 - 4 + 2 = 0$ ✓.

Answer $p = 2, q = - 2$

Example 81 · Advanced

Someone wrote

8 x^{3} - 27 = (2 x - 3) (4 x^{2} - 6 x + 9)

. Check by substitution whether it is right.

Solution, step by step

1Put $x = 1$ : left $8 - 27 = - 19$ .
2Right: $(2 - 3) (4 - 6 + 9) = (- 1) (7) = - 7$ .
3 $- 19 \neq = - 7$ : the answer is wrong. In a difference of cubes the second bracket is $a^{2} + ab + b^{2}$ , so the right form is $(2 x - 3) (4 x^{2} + 6 x + 9)$ : at $x = 1$ , $(- 1) (19) = - 19$ ✓.

Answer wrong; right: $(2 x - 3) (4 x^{2} + 6 x + 9)$

This two-minute check catches mistakes in any factorisation or expansion you do.

Practice

BasicTry it yourself first

Which of these statements is not an identity?

BasicTry it yourself first

Which of these statements is not an identity?

StandardTry it yourself first

Which of these statements is not an identity?

StandardTry it yourself first

Which of these statements is not an identity?

StandardTry it yourself first

What is

(a - 2 b)^{2} + 8 ab

equal to?

StandardTry it yourself first

What is

(a - 3 b)^{2} + 12 ab

equal to?

StandardTry it yourself first

(x + 2) (x^{2} + p x + q) = x^{3} + 5 x^{2} + 5 x - 2

holds for all

x

. Find

p + q

AdvancedTry it yourself first

(x - 3) (x^{2} + p x + q) = x^{3} - x^{2} - 2 x - 12

holds for all

x

. Find

p + q

AdvancedTry it yourself first

Which of these statements is not an identity?

AdvancedTry it yourself first

What is

(a - 5 b)^{2} + 20 ab

equal to?

Summary, mixed practice, revision

All methods in one place

One table of all identities, exam-style mixed questions and revision cards.

Summary of all methods

Which identity when

One page to choose the right identity as soon as you read a question.

Shape of the question	Identity / move
a number squared, near 100 or ending in 5	$(a \pm b)^{2} = a^{2} \pm 2 ab + b^{2}$ ; $(10 k + 5)^{2} = 100 k (k + 1) + 25$
product of two numbers at equal distances	$(a + b) (a - b) = a^{2} - b^{2}$
$a + b, ab$ given; $a^{2} + b^{2}$ or $\frac{1}{a} + \frac{1}{b}$	$a^{2} + b^{2} = S^{2} - 2 P$ ; $(a - b)^{2} = S^{2} - 4 P$ ; $\frac{S}{P}$
a cubed bracket, or $a^{3} + b^{3}$ asked	$(a \pm b)^{3}$ ; $a^{3} + b^{3} = S^{3} - 3 PS$ ; $a^{3} - b^{3} = D^{3} + 3 P D$
sum/difference of cubes, in a fraction	$a^{3} \pm b^{3} = (a \pm b) (a^{2} \mp ab + b^{2})$
a big $x$ , $x^{3} + 3 m x^{2} + 3 m^{2} x$	$(x + m)^{3} - m^{3}$
one coefficient repeated, equal to $x \pm 1$	write it as $x \pm 1$ and cancel the pairs
three letters, two of $S, q, T$ given	$S^{2} = T + 2 q$
three close numbers / $E$ / a ratio	$a^{3} + b^{3} + c^{3} - 3 ab c = \frac{1}{2} (a + b + c) \sum (a - b)^{2}$
sum zero, $a + b = c$ , a cycle of differences	$a^{3} + b^{3} + c^{3} = 3 ab c$ (carry the sign of a subtracted letter)
symmetric expressions in three letters	$p^{2} - 2 q, p^{3} - 3 pq + 3 r, pq - r, \frac{q}{r}$
$a^{4} + a^{2} b^{2} + b^{4}$ , $a^{4} + 4 b^{4}$ , $x^{4} + y^{4}$	add and subtract a square; square twice
one equation, several letters, sum of squares zero	each square separately $0$
factorise	common factor → shape → identity → check
remainder / is a factor / find $k$	put in the root: $f (r)$
least / greatest	complete the square; $A + B \geq 2 A B$
which option is right / an identity?	test with different, non-zero values

Here: $S = a + b$ , $D = a - b$ , $P = ab$ , $q = ab + b c + c a$ , $T = a^{2} + b^{2} + c^{2}$ , $p = a + b + c$ , $r = ab c$ , $E = a^{2} + b^{2} + c^{2} - ab - b c - c a$ , $f (r)$ = the value of the polynomial at $x = r$ .

First lookAfter reading a question ask three things: “how many letters?” “what is given, what is asked?” “is some sum zero, or some square equal to zero?” They lead you to the right row.

Mixed practice — set 1

Basic to standard

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

These questions are not in topic order — just like the exam. For each, first decide which identity fits.

BasicTry it yourself first

Find the value of

29 8^{2}

BasicTry it yourself first

Find

81 \times 89

using the identity.

BasicTry it yourself first

a + b = 13

and

ab = 40

, find

a^{2} + b^{2}

BasicTry it yourself first

Expand

(3 x - y)^{3}

BasicTry it yourself first

a + b + c = 14

and

a^{2} + b^{2} + c^{2} = 70

, find

ab + b c + c a

BasicTry it yourself first

For

a = 73, b = 69, c = 66

, find

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

BasicTry it yourself first

For

a = 9, b = - 6, c = - 3

, find

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

BasicTry it yourself first

Factorise

x^{2} + 10 x + 24

BasicTry it yourself first

What is the remainder when

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

is divided by

x + 2

BasicTry it yourself first

Find the least value of

y = x^{2} - 10 x + 31

BasicTry it yourself first

For real

x, y, z

(x - 5)^{2} + (y + 3)^{2} + (z - 1)^{2} = 0

. Find the value of

x + y + z

StandardTry it yourself first

Expand

(6 x + 5 y)^{2}

StandardTry it yourself first

Factorise

27 x^{3} - 64 y^{3}

StandardTry it yourself first

a + b = 9

and

ab = 20

, find

a^{3} + b^{3}

StandardTry it yourself first

a + b + c = 14

ab + b c + c a = 59

ab c = 70

, find

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

StandardTry it yourself first

Which of these statements is not an identity?

Mixed practice — set 2

Standard to advanced

Mixed questions from the hard end of the exam.

StandardTry it yourself first

For real

x, y

2 x^{2} + 6 x y + 9 y^{2} + 18 y + 18 = 0

. Find the value of

x + y

AdvancedTry it yourself first

For

x = 152

, find the value of

x^{3} - 6 x^{2} + 12 x

AdvancedTry it yourself first

a + b = 11

and

ab = 24

, find

a^{4} + b^{4}

AdvancedTry it yourself first

a + b = 7

a - b = 5

and

a^{2} + b^{2} = 37

, find

a^{4} - b^{4}

AdvancedTry it yourself first

a + b = 9

and

a^{3} + b^{3} = 351

, find

ab

AdvancedTry it yourself first

a + b + c = 17

a^{2} + b^{2} + c^{2} = 109

and

a^{3} + b^{3} + c^{3} = 755

, find

ab c

AdvancedTry it yourself first

For

a = 63, b = 60, c = 59

, find

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

AdvancedTry it yourself first

For

x = 17, y = 3, z = 7

, find

(x - 9)^{3} + (y - 9)^{3} + (z - 9)^{3}

AdvancedTry it yourself first

a + b + c = 15

ab + b c + c a = 68

ab c = 96

, find

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

AdvancedTry it yourself first

ab + b c + c a = 54

and

ab c = 72

, find

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

AdvancedTry it yourself first

a^{2} + ab + b^{2} = 91

and

a^{4} + a^{2} b^{2} + b^{4} = 2, 821

, find

ab

AdvancedTry it yourself first

For real

x, y

with

x + y = 9

and

x^{4} + y^{4} = 2, 417

, find

x y

AdvancedTry it yourself first

For real

x, y, z

(x + y - 11)^{2} + (y + z - 8)^{2} + (z + x - 13)^{2} = 0

. Find

x + y + z

AdvancedTry it yourself first

Factorise

8 a^{3} + b^{3} + 27 c^{3} - 18 ab c

AdvancedTry it yourself first

(x - 2)

and

(x + 3)

are both factors of

x^{3} + a x^{2} + b x + 6

. Find

a + b

AdvancedTry it yourself first

Find the remainder when

x^{25} + x^{49} - 20

is divided by

x + 1

AdvancedTry it yourself first

For

x > 0

, find the least value of

6 x + \frac{24}{x}

AdvancedTry it yourself first

A wire 84 m long is bent into a rectangle. Find the greatest possible area in square metres.

AdvancedTry it yourself first

What is

(a - 4 b)^{2} + 16 ab

equal to?

AdvancedTry it yourself first

Find the least value of

y = 4 x^{2} - 24 x + 41

AdvancedTry it yourself first

Factorise

x^{4} + 16 x^{2} + 256

Revision — cards and traps

Flip and recall

Read the front, think, then flip.

Trap 1Writing $(a + b)^{2}$ as $a^{2} + b^{2}$ . Do not forget the middle

2 ab

; test: at

a = 2, b = 1

9 \neq = 5

Trap 2Not squaring or cubing the number part.

(3 x)^{2} = 9 x^{2}

(2 x)^{3} = 8 x^{3}

Trap 3Wrong sign in the second bracket of $a^{3} \pm b^{3}$ .

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

for the sum,

a^{2} + ab + b^{2}

for the difference; the last term is always

+ b^{2}

Trap 4Making every sign minus in $(a - b)^{3}$ . The signs alternate

+, -, +, -

Trap 5Taking only one sign of a square root.

(a + b)^{2} = 100

gives

a + b = \pm 10

; only

+ 10

if positivity is stated.

Trap 6Jumping from $a^{3} + b^{3} + c^{3} = 3 ab c$ to $a = b = c$ . Only when

a + b + c \neq = 0

(for example all positive); otherwise

a + b + c = 0

is also possible.

Trap 7Keeping both roots of a quadratic. For real

x, y

(x - y)^{2} \geq 0

can remove one root.

Trap 8Trusting one pair of test values. At

a = 0

b = 0

a = b

many wrong options look right; try a second pair.