x + 1/x Type Expressions

SSC CGL · CHSL · CPO — method by method

3 Part48 worked examples147 practice questions

Exam questions often give one small fact, such as $x + \frac{1}{x} = 7$ , and ask for something heavy, such as $x^{4} + \frac{1}{x ^{4}}$ or $x^{3} - \frac{1}{x ^{3}}$ . You never need to find x itself. The secret is that $x$ and $\frac{1}{x}$ multiply to exactly $1$ . So when we square or cube, the middle terms that would contain $x$ turn into plain numbers.

∑ Formula

x \times \frac{1}{x} = 1

The foundation of the whole chapter

$x$: any number except zero
$\frac{1}{x}$: the reciprocal (upside-down) of x

The chapter has two parts. The ladder teaches how $x + \frac{1}{x}$ builds $x^{2} + \frac{1}{x ^{2}}$ , then $x^{3} + \frac{1}{x ^{3}}$ and the higher powers, and how to walk back down. Reaching the right form teaches what to do when the question does not hand you $x + \frac{1}{x}$ directly (an equation, a fraction, a scaled pair or a shifted variable is given). Special values ( $x + \frac{1}{x} = 1, - 1, 3, \dots$ ), huge powers, three-variable questions and minimum values are in the next chapter, “x + 1/x: Cycles, Limits and Several Variables”.

First questionIn every question first ask: “Is the pair

x

and

\frac{1}{x}

, with product 1?” If yes, never touch x itself; climb the ladder.

The ladder: higher powers

Square · cube · back · multiply · the minus family

From x + 1/x to x² + 1/x², x³ + 1/x³ …, and the x − 1/x family.

These five methods are the backbone of the chapter. Read them in order: each stands on the one before.

The square: x² + 1/x²

a^{2} - 2

and

b^{2} + 2

The first rung: square $x + \frac{1}{x}$ or $x - \frac{1}{x}$ .

The idea

Square $x + \frac{1}{x}$ : $(x + \frac{1}{x})^{2} = x^{2} + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x ^{2}}$ . In the middle term $x$ and $\frac{1}{x}$ cancel, leaving a plain $2$ . So to get $x^{2} + \frac{1}{x ^{2}}$ we simply subtract that $2$ . The picture below shows it: the big square has side $x + \frac{1}{x}$ and four pieces: $x^{2}$ , two rectangles of area $x \cdot \frac{1}{x} = 1$ , and the small square $\frac{1}{x ^{2}}$ .

◭ Diagram

Pieces of the big square (side

x + \frac{1}{x}

): the two saffron rectangles are 1 each, that is the “

- 2

”. (The picture uses

x = 2

, but it holds for every

x

Squaring $x - \frac{1}{x}$ gives a middle term $- 2$ , so this time we add. The two squares always differ by $4$ ; that is the bridge for going between $x + \frac{1}{x}$ and $x - \frac{1}{x}$ .

∑ Formula

x + \frac{1}{x} = a \Rightarrow x^{2} + \frac{1}{x ^{2}} = a^{2} - 2

The plus form

$a$: the given value of $x + \frac{1}{x}$
$x^{2} + \frac{1}{x ^{2}}$: the value we want

∑ Formula

x - \frac{1}{x} = b \Rightarrow x^{2} + \frac{1}{x ^{2}} = b^{2} + 2

The minus form

$b$: the given value of $x - \frac{1}{x}$

∑ Formula

(x + \frac{1}{x})^{2} - (x - \frac{1}{x})^{2} = 4

The bridge from one to the other

$(x + \frac{1}{x})^{2}$: square of the sum
$(x - \frac{1}{x})^{2}$: square of the difference

$x + \frac{1}{x}$	$x^{2} + \frac{1}{x ^{2}}$
$2$	$2$
$3$	$7$
$4$	$14$
$5$	$23$
$6$	$34$
$7$	$47$
$8$	$62$
$9$	$79$
$10$	$98$
$11$	$119$
$12$	$142$

$x - \frac{1}{x}$	$x^{2} + \frac{1}{x ^{2}}$
$1$	$3$
$2$	$6$
$3$	$11$
$4$	$18$
$5$	$27$
$6$	$38$
$7$	$51$
$8$	$66$
$9$	$83$
$10$	$102$

Here: in both tables the left column is the given value and the right column is $x^{2} + \frac{1}{x ^{2}}$ . If you learn to recognise these numbers (7, 14, 23, 34, 47, 62 and 3, 6, 11, 18, 27, 38), later questions open at once.

Spot itSpot it: the question gives the sum or difference of

x

and

\frac{1}{x}

and asks the

x^{2}

version. Square.

Steps

See whether the pair is $x + \frac{1}{x}$ or $x - \frac{1}{x}$ .
Square both sides.
The middle term is the number $\pm 2$ : subtract it for a plus pair, add it for a minus pair.

Example 1 · Basic

x + \frac{1}{x} = 8

, find

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

Solution, step by step

1We are given $x + \frac{1}{x}$ and want the $x^{2}$ version, so we square both sides.
2 $(x + \frac{1}{x})^{2} = x^{2} + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x ^{2}} = x^{2} + 2 + \frac{1}{x ^{2}}$ , because $x \cdot \frac{1}{x} = 1$ .
3The left side is $8^{2} = 64$ , so $x^{2} + \frac{1}{x ^{2}} + 2 = 64$ .
4Moving the $2$ across gives $x^{2} + \frac{1}{x ^{2}} = 64 - 2 = 62$ .

Answer 62

Direct formula:

a^{2} - 2 = 64 - 2 = 62

Example 2 · Standard

x - \frac{1}{x} = 6

, find

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

Solution, step by step

1This time the pair is a difference. Squaring gives $(x - \frac{1}{x})^{2} = x^{2} - 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x ^{2}} = x^{2} - 2 + \frac{1}{x ^{2}}$ .
2The left side is $6^{2} = 36$ : $x^{2} + \frac{1}{x ^{2}} - 2 = 36$ .
3Now the $- 2$ crosses over as $+ 2$ : $x^{2} + \frac{1}{x ^{2}} = 36 + 2 = 38$ .

Answer 38

Trap: a minus pair uses “

+ 2

”, not “

- 2

”. The square of

x - \frac{1}{x}

adds instead of subtracting.

Example 3 · Standard

x + \frac{1}{x} = 6

and

x > 1

, find

x - \frac{1}{x}

Solution, step by step

1Use the bridge: $(x - \frac{1}{x})^{2} = (x + \frac{1}{x})^{2} - 4$ , because the two squares sit $+ 2$ and $- 2$ away from $x^{2} + \frac{1}{x ^{2}}$ .
2 $(x - \frac{1}{x})^{2} = 36 - 4 = 32$ .
3As $x > 1$ , we have $x > \frac{1}{x}$ and $x - \frac{1}{x} > 0$ . So $x - \frac{1}{x} = 32 = 42$ .

Answer 4√2

x < 1

were given the answer would be

- 42

. Read the sign condition when you take a square root.

Example 4 · Advanced

x + \frac{1}{x} = \frac{13}{6}

, find

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

and check it by finding

x

Solution, step by step

1The formula $a^{2} - 2$ with $a = \frac{13}{6}$ gives $(\frac{13}{6})^{2} - 2 = \frac{169}{36} - \frac{72}{36} = \frac{97}{36}$ .
2Check: $x + \frac{1}{x} = \frac{13}{6}$ gives $6 x^{2} - 13 x + 6 = 0$ , i.e. $(2 x - 3) (3 x - 2) = 0$ , so $x = \frac{3}{2}$ or $\frac{2}{3}$ .
3Taking $x = \frac{3}{2}$ : $x^{2} + \frac{1}{x ^{2}} = \frac{9}{4} + \frac{4}{9} = \frac{81 + 16}{36} = \frac{97}{36}$ . Both ways agree.

Answer 97/36

Swapping

x

and

\frac{1}{x}

changes nothing, so both roots give the same answer.

Example 5 · Advanced

x - \frac{1}{x} = 7

, find

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

and

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

Solution, step by step

1 $(7)^{2} = 7$ . So $x^{2} + \frac{1}{x ^{2}} = 7 + 2 = 9$ .
2 $(x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x ^{2}} + 2 = 9 + 2 = 11$ .
3Check with the bridge: $7 + 4 = 11$ , correct.

Answer 9 ; 11

Squaring a square-root value removes the root, so the answers are whole numbers.

Practice

BasicTry it yourself first

x + \frac{1}{x} = 11

, then

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

equals:

BasicTry it yourself first

x - \frac{1}{x} = 9

, then

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

equals:

StandardTry it yourself first

x + \frac{1}{x} = - 7

, then

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

equals:

StandardTry it yourself first

x - \frac{1}{x} = 7

, then

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

equals:

StandardTry it yourself first

x + \frac{1}{x} = \frac{10}{3}

, then

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

equals:

StandardTry it yourself first

x + \frac{1}{x} = 11

, then

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

equals:

StandardTry it yourself first

x + \frac{1}{x} = 10

and

x > 1

, then

x - \frac{1}{x}

equals:

AdvancedTry it yourself first

x - \frac{1}{x} = 4

and

x > 0

, then

x + \frac{1}{x}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 9

, then

(x - \frac{1}{x})^{2}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 25

, then

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

equals:

Going back: from the square to x ± 1/x

a + 2

and

a - 2

When $x^{2} + \frac{1}{x ^{2}}$ or $x^{4} + \frac{1}{x ^{4}}$ is given and you want $x + \frac{1}{x}$ .

The idea

Read the previous formula backwards. Squaring subtracted $2$ , so going back we add $2$ and take a square root: $(x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x ^{2}} + 2$ and $(x - \frac{1}{x})^{2} = x^{2} + \frac{1}{x ^{2}} - 2$ . A square root brings two signs ( $\pm$ ); it is the condition in the question ( $x > 0$ , $x > 1$ …) that tells you which to choose.

∑ Formula

x^{2} + \frac{1}{x ^{2}} = a \Rightarrow x + \frac{1}{x} = \pm a + 2, x - \frac{1}{x} = \pm a - 2

Both square roots

$a$: the given value of $x^{2} + \frac{1}{x ^{2}}$ (always $\geq 2$ )
$\pm$: two possible signs; pick by the condition

For $x^{4} + \frac{1}{x ^{4}}$ do the same job twice: first reach $x^{2} + \frac{1}{x ^{2}}$ , then $x \pm \frac{1}{x}$ . $x^{2} + \frac{1}{x ^{2}}$ is always positive, so take only the $+$ root there.

$x^{2} + \frac{1}{x ^{2}}$	$x + \frac{1}{x}$ ( $x > 0$ )	$x^{4} + \frac{1}{x ^{4}}$
$7$	$3$	$47$
$14$	$4$	$194$
$23$	$5$	$527$
$34$	$6$	$1154$
$47$	$7$	$2207$
$62$	$8$	$3842$
$79$	$9$	$6239$
$98$	$10$	$9602$

Here: the middle column is the $x + \frac{1}{x}$ that produces the left and right values. If you know these triples, seeing $2207$ immediately tells you $47$ and $7$ are behind it.

Spot itSpot it: the given value is “a square

- 2

”:

7, 14, 23, 34, 47, 62, 79, 98, 119, \dots

(square of

x + \frac{1}{x}

minus 2) or “a square

+ 2

” (

3, 6, 11, 18, 27, 38, \dots

). Trap:

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

can never be negative, nor less than

2

Steps

Add $2$ to the given value (and/or subtract $2$ ).
Take the square root; if the number is not a perfect square, leave the root as it is.
Choose the sign from the condition ( $x > 0$ ⟹ $x + \frac{1}{x} > 0$ ; $x > 1$ ⟹ $x - \frac{1}{x} > 0$ ).
For a fourth power, first come down to $x^{2}$ .

Example 6 · Basic

x^{2} + \frac{1}{x ^{2}} = 51

and

x > 0

, find

x + \frac{1}{x}

Solution, step by step

1The square of $x + \frac{1}{x}$ is $2$ more than $x^{2} + \frac{1}{x ^{2}}$ , so $(x + \frac{1}{x})^{2} = 51 + 2 = 53$ .
2 $53$ is not a perfect square, so the root stays as it is: $x + \frac{1}{x} = \pm 53$ .
3As $x > 0$ , both $x$ and $\frac{1}{x}$ are positive and so is the sum: $x + \frac{1}{x} = 53$ .

Answer √53

You do not always get a perfect square; leaving the root is also a correct answer.

Example 7 · Standard

x^{4} + \frac{1}{x ^{4}} = 1154

and

x > 1

, find

x + \frac{1}{x}

and

x - \frac{1}{x}

Solution, step by step

1Come down from $x^{4} + \frac{1}{x ^{4}}$ to $x^{2} + \frac{1}{x ^{2}}$ : $(x^{2} + \frac{1}{x ^{2}})^{2} = 1154 + 2 = 1156 = 3 4^{2}$ , and $x^{2} + \frac{1}{x ^{2}} > 0$ , so $x^{2} + \frac{1}{x ^{2}} = 34$ .
2Sum: $(x + \frac{1}{x})^{2} = 34 + 2 = 36$ , and $x > 0$ , so $x + \frac{1}{x} = 6$ .
3Difference: $(x - \frac{1}{x})^{2} = 34 - 2 = 32$ , and $x > 1$ , so $x - \frac{1}{x} = 32 = 42$ .

Answer 6 ; 4√2

Check the pair together:

6^{2} - (42)^{2} = 36 - 32 = 4

✓.

Example 8 · Advanced

x^{2} + \frac{1}{x ^{2}} = 46

and

x > 1

, find

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

Solution, step by step

1Here the base is $x^{2}$ ; $x^{2} \cdot \frac{1}{x ^{2}} = 1$ , so the same bridge works: $(x^{2} - \frac{1}{x ^{2}})^{2} = (x^{2} + \frac{1}{x ^{2}})^{2} - 4$ .
2 $= 4 6^{2} - 4 = 2116 - 4 = 2112$ .
3 $2112 = 64 \times 33$ , so $2112 = 833$ . $x > 1$ ⟹ $x^{2} > \frac{1}{x ^{2}}$ , so take the positive root: $833$ .

Answer 8√33

Alternative:

x^{2} - \frac{1}{x ^{2}} = (x + \frac{1}{x}) (x - \frac{1}{x}) = 48 \cdot 44 = 2112

, the same.

Practice

BasicTry it yourself first

x^{2} + \frac{1}{x ^{2}} = 98

and

x > 0

, then

x + \frac{1}{x}

equals:

BasicTry it yourself first

x^{2} + \frac{1}{x ^{2}} = 79

and

x > 0

, then

x + \frac{1}{x}

equals:

StandardTry it yourself first

x^{4} + \frac{1}{x ^{4}} = 194

, then

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

equals:

StandardTry it yourself first

x^{4} + \frac{1}{x ^{4}} = 2207

and

x > 0

, then

x + \frac{1}{x}

equals:

StandardTry it yourself first

x^{2} + \frac{1}{x ^{2}} = 34

and

x > 1

, then

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

equals:

StandardTry it yourself first

x^{2} + \frac{1}{x ^{2}} = 30

and

x > 0

, then

x + \frac{1}{x}

equals:

StandardTry it yourself first

Which of these can NOT be the value of

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

for any real

x

AdvancedTry it yourself first

x^{4} + \frac{1}{x ^{4}} = 6239

and

x > 0

, then

x + \frac{1}{x}

equals:

AdvancedTry it yourself first

x^{2} + \frac{1}{x ^{2}} = 47

and

x > 1

, then

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

equals:

AdvancedTry it yourself first

x^{2} + \frac{1}{x ^{2}} = 83

and

x > 1

, then

x - \frac{1}{x}

equals:

The cube: x³ + 1/x³

a^{3} - 3 a

and

b^{3} + 3 b

The second rung: cube the pair.

The idea

Cube the pair: $(x + \frac{1}{x})^{3} = x^{3} + 3 x^{2} \cdot \frac{1}{x} + 3 x \cdot \frac{1}{x ^{2}} + \frac{1}{x ^{3}}$ . In both middle terms the $x$ cancels and what is left is $3 x + \frac{3}{x} = 3 (x + \frac{1}{x})$ , three times the pair itself. So subtract three times the pair from the cube and $x^{3} + \frac{1}{x ^{3}}$ remains.

◭ Diagram

For

x + \frac{1}{x} = 4

: the cube

64

splits into

x^{3} + \frac{1}{x ^{3}} = 52

and

3 (x + \frac{1}{x}) = 12

∑ Formula

x + \frac{1}{x} = a \Rightarrow x^{3} + \frac{1}{x ^{3}} = a^{3} - 3 a

The plus cube

$a$: the value of $x + \frac{1}{x}$
$a^{3}$: cube of the pair
$3 a$: the middle terms together: three times the pair

∑ Formula

x - \frac{1}{x} = b \Rightarrow x^{3} - \frac{1}{x ^{3}} = b^{3} + 3 b

The minus cube

$b$: the value of $x - \frac{1}{x}$

Another view: $x^{3} + \frac{1}{x ^{3}} = (x + \frac{1}{x}) (x^{2} - 1 + \frac{1}{x ^{2}})$ (sum-of-cubes factorisation). Put $x^{2} + \frac{1}{x ^{2}} = a^{2} - 2$ in the bracket and you get $a (a^{2} - 2 - 1) = a^{3} - 3 a$ : the same formula by a second road.

$x + \frac{1}{x}$	$x^{3} + \frac{1}{x ^{3}}$
$2$	$2$
$3$	$18$
$4$	$52$
$5$	$110$
$6$	$198$
$7$	$322$
$8$	$488$
$9$	$702$
$10$	$970$

$x - \frac{1}{x}$	$x^{3} - \frac{1}{x ^{3}}$
$1$	$4$
$2$	$14$
$3$	$36$
$4$	$76$
$5$	$140$
$6$	$234$
$7$	$364$
$8$	$536$
$9$	$756$

Here: the left column is the given value, the right column the cube version. You need not memorise these pairs, but having seen a few ( $4 \to 52$ , $5 \to 110$ , $6 \to 198$ , $7 \to 322$ ) makes the options easy to recognise.

ReverseReverse question: if

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

is given and you want

x + \frac{1}{x}

, solve

a^{3} - 3 a = c

from the table or by estimating (

a \approx 3 c

); in questions

a

is almost always a whole number.

Steps

Get the value of $x + \frac{1}{x}$ (or $x - \frac{1}{x}$ ) ready; if only the $x^{2}$ value is given, first come back by a square root.
Apply the cube formula: $- 3 a$ for a plus pair, $+ 3 b$ for a minus pair.
Check signs: if $a$ is negative, both $a^{3}$ and $3 a$ are negative.

Example 9 · Basic

x + \frac{1}{x} = 6

, find

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

Solution, step by step

1Cube the pair: $(x + \frac{1}{x})^{3} = x^{3} + \frac{1}{x ^{3}} + 3 (x + \frac{1}{x})$ .
2On the left $6^{3} = 216$ , and three times the pair is $3 \times 6 = 18$ .
3So $x^{3} + \frac{1}{x ^{3}} = 216 - 18 = 198$ .

Answer 198

Trap: subtract

3 a

, not just

3

Example 10 · Standard

x - \frac{1}{x} = 7

, find

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

Solution, step by step

1 $(x - \frac{1}{x})^{3} = x^{3} - 3 x + \frac{3}{x} - \frac{1}{x ^{3}} = (x^{3} - \frac{1}{x ^{3}}) - 3 (x - \frac{1}{x})$ , since $- 3 x + \frac{3}{x} = - 3 (x - \frac{1}{x})$ .
2So $x^{3} - \frac{1}{x ^{3}} = (x - \frac{1}{x})^{3} + 3 (x - \frac{1}{x}) = 343 + 21 = 364$ .

Answer 364

The minus form carries

+ 3 b

: both signs flip compared with the plus form.

Example 11 · Standard

x^{2} + \frac{1}{x ^{2}} = 47

and

x > 0

, find

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

Solution, step by step

1The cube formula needs $x + \frac{1}{x}$ , so first go back: $(x + \frac{1}{x})^{2} = 47 + 2 = 49$ .
2As $x > 0$ , $x + \frac{1}{x} = 7$ .
3 $x^{3} + \frac{1}{x ^{3}} = 7^{3} - 3 \times 7 = 343 - 21 = 322$ .

Answer 322

You cannot skip the middle rung: you do not go from

x^{2}

straight to

x^{3}

Example 12 · Advanced

x^{3} + \frac{1}{x ^{3}} = 488

and

x > 0

, find

x + \frac{1}{x}

Solution, step by step

1Let $a = x + \frac{1}{x}$ . Then $a^{3} - 3 a = 488$ , and since $x > 0$ , $a \geq 2$ .
2 $a^{3}$ is about $488$ , between $7^{3} = 343$ and $8^{3} = 512$ , so try $a = 8$ : $512 - 24 = 488$ ✓.
3For $a \geq 2$ , $a^{3} - 3 a$ keeps increasing, so there is no second solution: $x + \frac{1}{x} = 8$ .

Answer 8

Fine check:

a = 7

gives

322

and

a = 9

gives

702

;

488

lies between, so

8

is right.

Practice

BasicTry it yourself first

x + \frac{1}{x} = 9

, then

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

equals:

BasicTry it yourself first

x - \frac{1}{x} = 8

, then

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

equals:

StandardTry it yourself first

x + \frac{1}{x} = - 6

, then

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

equals:

StandardTry it yourself first

x^{2} + \frac{1}{x ^{2}} = 98

and

x > 0

, then

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

equals:

StandardTry it yourself first

x^{3} + \frac{1}{x ^{3}} = 702

and

x > 0

, then

x + \frac{1}{x}

equals:

StandardTry it yourself first

x - \frac{1}{x} = 6

, then

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

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 23

, then

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

equals:

AdvancedTry it yourself first

x - \frac{1}{x} = 6

, then

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

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = \frac{5}{2}

, then

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

equals:

AdvancedTry it yourself first

x^{3} + \frac{1}{x ^{3}} = 1298

and

x > 0

, then

x + \frac{1}{x}

equals:

Higher powers: multiply two rungs

a_{m + n} = a_{m} a_{n} - a_{m - n}

The rule that reaches $x^{4}, x^{5}, x^{6}, x^{7}, \dots$ and the shortest road there.

The idea

For convenience write $a_{n}$ for $x^{n} + \frac{1}{x ^{n}}$ . Now multiply two rungs: $(x^{m} + \frac{1}{x ^{m}}) (x^{n} + \frac{1}{x ^{n}}) = x^{m + n} + x^{m - n} + x^{n - m} + \frac{1}{x ^{m + n}}$ . The first and last terms together make $a_{m + n}$ ; the two middle terms together make $x^{m - n} + \frac{1}{x ^{m - n}} = a_{m - n}$ . So product = upper rung + the rung of the difference of the powers, and upper rung = product − difference rung.

∑ Formula

a_{m + n} = a_{m} a_{n} - a_{m - n} (a_{0} = 2, a_{1} = x + \frac{1}{x})

The upper rung

$a_{n}$: $x^{n} + \frac{1}{x ^{n}}$
$m, n$: two powers, $m \geq n$
$a_{m - n}$: the rung of the difference; if $m = n$ it is $a_{0} = x^{0} + \frac{1}{x ^{0}} = 2$

Want	Shortest way
$a_{4}$	$a_{2} \cdot a_{2} - 2$
$a_{5}$	$a_{2} \cdot a_{3} - a_{1}$
$a_{6}$	$a_{3} \cdot a_{3} - 2$
$a_{7}$	$a_{3} \cdot a_{4} - a_{1}$
$a_{8}$	$a_{4} \cdot a_{4} - 2$
$a_{9}$	$a_{4} \cdot a_{5} - a_{1}$
$a_{10}$	$a_{5} \cdot a_{5} - 2$
any $a_{n + 1}$	$a_{1} \cdot a_{n} - a_{n - 1}$

Here: the $a_{k}$ on the right are the values already found by the earlier methods; $a_{1} = x + \frac{1}{x}$ . The last row is the rule that always works (one rung up each time); the other rows are shorter roads.

◭ Diagram

The ladder for

x + \frac{1}{x} = 6

: every new box is “

6 \times

the last

-

the one before”; the curved arrows below show the fourth box built from the third and second.

$n$	$x^{n} + \frac{1}{x ^{n}}$ (for $x + \frac{1}{x} = 3$ )
$0$	$2$
$1$	$3$
$2$	$7$
$3$	$18$
$4$	$47$
$5$	$123$
$6$	$322$
$7$	$843$
$8$	$2207$

Here: $n$ is the power and the right column is its value; the case $x + \frac{1}{x} = 3$ only shows that every row is made from the two rows above ( $7 = 3 \cdot 3 - 2$ , $18 = 3 \cdot 7 - 3$ , $47 = 3 \cdot 18 - 7$ , …).

TrickLast-digit trick: if the options all end in different digits, carry only the last digit of every rung. No need for the full big multiplications.

Steps

Write $a_{1} = x + \frac{1}{x}$ and get $a_{2} = a_{1}^{2} - 2$ .
Split the target power as a sum of two smaller powers ( $m + n$ ) so that the rung for the difference $m - n$ is already known.
Compute $a_{m} \cdot a_{n} - a_{m - n}$ . If $m = n$ subtract $2$ .
Check the answer against the last digits of the options.

Example 13 · Basic

x + \frac{1}{x} = 6

, find

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

Solution, step by step

1First the $x^{2}$ rung, then square it. Square both sides: $x^{2} + \frac{1}{x ^{2}} = (x + \frac{1}{x})^{2} - 2 = 6^{2} - 2 = 34$ .
2Now square the $x^{2}$ value: $x^{4} + \frac{1}{x ^{4}} = (x^{2} + \frac{1}{x ^{2}})^{2} - 2 = 3 4^{2} - 2 = 1154$ .
3So $x^{4} + \frac{1}{x ^{4}} = 1154$ .

Answer 1154

x^{4}

is really

(x^{2})^{2}

, so the base here is

x^{2}

and the square formula (

- 2

) applies.

Example 14 · Standard

x + \frac{1}{x} = 7

, find

x^{5} + \frac{1}{x ^{5}}

Solution, step by step

1 $5 = 2 + 3$ , so we need the $x^{2}$ rung and the $x^{3}$ rung. $a_{2} = 7^{2} - 2 = 47$ .
2 $a_{3} = 7^{3} - 3 \times 7 = 343 - 21 = 322$ .
3Multiply: $(x^{2} + \frac{1}{x ^{2}}) (x^{3} + \frac{1}{x ^{3}}) = x^{5} + x^{- 1} + x + \frac{1}{x ^{5}} = a_{5} + a_{1}$ (difference of powers $3 - 2 = 1$ ).
4So $a_{5} = 47 \times 322 - 7 = 15134 - 7 = 15127$ .

Answer 15127

Check: last digits

7 \times 2 - 7 \to 7

✓.

Example 15 · Standard

x + \frac{1}{x} = 8

, find

x^{6} + \frac{1}{x ^{6}}

by two different roads.

Solution, step by step

1 $a_{2} = 8^{2} - 2 = 62$ and $a_{3} = 8^{3} - 3 \times 8 = 512 - 24 = 488$ .
2Road 1 ( $6 = 3 + 3$ ): $a_{6} = a_{3} \cdot a_{3} - 2 = 48 8^{2} - 2 = 238144 - 2 = 238142$ .
3Road 2 ( $6 = 4 + 2$ ): first $a_{4} = 6 2^{2} - 2 = 3842$ ; then $a_{6} = a_{4} \cdot a_{2} - a_{4 - 2} = 3842 \times 62 - 62 = 238204 - 62 = 238142$ .
4Both roads agree: $238142$ .

Answer 238142

In the exam take the road with the easier multiplication; Road 1 needs only one squaring.

Example 16 · Advanced

x + \frac{1}{x} = 9

, find

x^{7} + \frac{1}{x ^{7}}

and check it with the last digit.

Solution, step by step

1 $7 = 3 + 4$ , so $a_{7} = a_{3} \cdot a_{4} - a_{1}$ . First $a_{2} = 9^{2} - 2 = 79$ and $a_{3} = 9^{3} - 3 \times 9 = 729 - 27 = 702$ .
2 $a_{4} = a_{2}^{2} - 2 = 7 9^{2} - 2 = 6241 - 2 = 6239$ .
3Now $a_{7} = 702 \times 6239 - 9 = 4379778 - 9 = 4379769$ .
4Last-digit check: $a_{3} \to 2$ , $a_{4} \to 9$ , $a_{1} = 9$ , so $a_{7} \to 2 \times 9 - 9 = 9$ . The answer above also ends in $9$ ✓.

Answer 4379769

For a big multiplication the last-digit check is the cheapest safety net.

Practice

BasicTry it yourself first

x + \frac{1}{x} = 11

, then

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

equals:

BasicTry it yourself first

x + \frac{1}{x} = 6

, then

x^{5} + \frac{1}{x ^{5}}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 7

, then

x^{6} + \frac{1}{x ^{6}}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 4

, then

x^{7} + \frac{1}{x ^{7}}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 8

, which of these is

x^{5} + \frac{1}{x ^{5}}

StandardTry it yourself first

x + \frac{1}{x} = 4

, then

x^{8} + \frac{1}{x ^{8}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 5

, then

x^{5} + \frac{1}{x ^{5}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 7

, then

x^{5} + \frac{1}{x ^{5}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = \frac{5}{2}

, then

x^{6} + \frac{1}{x ^{6}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = - 5

, then

x^{5} + \frac{1}{x ^{5}}

equals:

The minus family: xⁿ − 1/xⁿ

Factorising and the second ladder

When the powers carry a minus sign, or $x - \frac{1}{x}$ is given.

The idea

Minus forms can be reached in two ways. Road one: factorise. The difference-of-squares rule $A^{2} - B^{2} = (A + B) (A - B)$ doubles the power: $x^{2} - \frac{1}{x ^{2}} = (x + \frac{1}{x}) (x - \frac{1}{x})$ and $x^{4} - \frac{1}{x ^{4}} = (x^{2} + \frac{1}{x ^{2}}) (x^{2} - \frac{1}{x ^{2}})$ . The cube rule $A^{3} - B^{3} = (A - B) (A^{2} + A B + B^{2})$ gives $x^{3} - \frac{1}{x ^{3}} = (x - \frac{1}{x}) (x^{2} + 1 + \frac{1}{x ^{2}})$ . Road two: a second ladder. If $x - \frac{1}{x} = b$ , the pair is $x$ and $- \frac{1}{x}$ , whose product is $- 1$ . The ladder is then $u_{n} = x^{n} + (- \frac{1}{x})^{n}$ : for odd $n$ this is $x^{n} - \frac{1}{x ^{n}}$ , for even $n$ it is $x^{n} + \frac{1}{x ^{n}}$ .

∑ Formula

x^{2 n} - \frac{1}{x ^{2 n}} = (x^{n} + \frac{1}{x ^{n}}) (x^{n} - \frac{1}{x ^{n}})

Difference of squares: doubles the power

$n$: any power ( $1, 2, 3, \dots$ )
$x^{2 n} - \frac{1}{x ^{2 n}}$: what we want

∑ Formula

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

Difference of cubes

$x - \frac{1}{x}$: first factor
$x^{2} + 1 + \frac{1}{x ^{2}}$: second factor, $= a^{2} - 1$ where $a = x + \frac{1}{x}$

∑ Formula

u_{n + 1} = b u_{n} + u_{n - 1}, u_{0} = 2, u_{1} = b

The second ladder (

b = x - \frac{1}{x}

)

$u_{n}$: $x^{n} + (- \frac{1}{x})^{n}$
$b$: the value of $x - \frac{1}{x}$

Note the “ $+ u_{n - 1}$ ” in this ladder instead of the “ $- a_{n - 1}$ ” of the first ladder, because the pair’s product is $- 1$ , not $+ 1$ . Below is the table for $b = 2$ .

$n$	$u_{n}$	Meaning
$0$	$2$	$x^{0} + \frac{1}{x ^{0}}$
$1$	$2$	$x^{1} - \frac{1}{x ^{1}}$
$2$	$6$	$x^{2} + \frac{1}{x ^{2}}$
$3$	$14$	$x^{3} - \frac{1}{x ^{3}}$
$4$	$34$	$x^{4} + \frac{1}{x ^{4}}$
$5$	$82$	$x^{5} - \frac{1}{x ^{5}}$
$6$	$198$	$x^{6} + \frac{1}{x ^{6}}$
$7$	$478$	$x^{7} - \frac{1}{x ^{7}}$

Here: with $x - \frac{1}{x} = 2$ , row $n$ gives $u_{n}$ and the last column says which expression it is. Each $u$ is twice the last plus the one before ( $6 = 2 \cdot 2 + 2$ , $14 = 2 \cdot 6 + 2$ , $34 = 2 \cdot 14 + 6$ ).

SignIf

x + \frac{1}{x}

is given and

x^{n} - \frac{1}{x ^{n}}

is asked, first get

x - \frac{1}{x} = \pm s^{2} - 4

, choose the sign from the condition (

x > 1

⟹

+

), then factorise. The sign is almost always settled in the question.

Steps

See whether the given pair is $x + \frac{1}{x}$ or $x - \frac{1}{x}$ , and whether the asked form has a plus or a minus.
If needed, build the other pair with $(x - \frac{1}{x})^{2} = (x + \frac{1}{x})^{2} - 4$ (sign from the condition).
For an even power use the difference of squares; for an odd power use the cube rule or the second ladder.

Example 17 · Basic

x + \frac{1}{x} = 6

and

x > 1

, find

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

Solution, step by step

1 $x^{2} - \frac{1}{x ^{2}}$ is a difference of squares, so split it into $(x + \frac{1}{x}) (x - \frac{1}{x})$ . We know $x + \frac{1}{x} = 6$ ; we need $x - \frac{1}{x}$ .
2 $(x - \frac{1}{x})^{2} = 36 - 4 = 32$ and $x > 1$ , so $x - \frac{1}{x} = 42$ .
3Multiply: $x^{2} - \frac{1}{x ^{2}} = 6 \times 42 = 242$ .

Answer 24√2

Check:

(242)^{2} = 1152

and

3 4^{2} - 4 = 1152

✓ (the

x^{2}

bridge from the earlier method).

Example 18 · Standard

x + \frac{1}{x} = 10

and

x > 1

, find

x^{4} - \frac{1}{x ^{4}}

Solution, step by step

1 $x^{4} - \frac{1}{x ^{4}} = (x^{2} + \frac{1}{x ^{2}}) (x^{2} - \frac{1}{x ^{2}})$ and $x^{2} - \frac{1}{x ^{2}} = (x + \frac{1}{x}) (x - \frac{1}{x})$ : the difference of squares twice.
2 $x - \frac{1}{x}$ : $100 - 4 = 96$ , $x > 1$ ⟹ $x - \frac{1}{x} = 96 = 46$ .
3 $x^{2} - \frac{1}{x ^{2}} = 10 \times 46 = 406$ and $x^{2} + \frac{1}{x ^{2}} = 100 - 2 = 98$ .
4So $x^{4} - \frac{1}{x ^{4}} = 98 \times 406 = 39206$ .

Answer 3920√6

Three factors one after another:

s \cdot d \cdot a_{2}

Example 19 · Standard

x - \frac{1}{x} = 2

, find

x^{5} - \frac{1}{x ^{5}}

Solution, step by step

1Second ladder: $u_{1} = 2$ . $u_{2} = x^{2} + \frac{1}{x ^{2}} = 2^{2} + 2 = 6$ .
2 $u_{3} = x^{3} - \frac{1}{x ^{3}} = 2^{3} + 3 \times 2 = 14$ .
3Multiply: $(x^{2} + \frac{1}{x ^{2}}) (x^{3} - \frac{1}{x ^{3}}) = x^{5} - x^{- 1} + x - x^{- 5} = (x^{5} - \frac{1}{x ^{5}}) + (x - \frac{1}{x})$ .
4So $x^{5} - \frac{1}{x ^{5}} = 6 \times 14 - 2 = 82$ .

Answer 82

Row

n = 5

of the table also gives

82

Example 20 · Advanced

x - \frac{1}{x} = 4

and

x > 0

, find

x^{4} - \frac{1}{x ^{4}}

Solution, step by step

1 $x^{4} - \frac{1}{x ^{4}} = (x^{2} + \frac{1}{x ^{2}}) (x + \frac{1}{x}) (x - \frac{1}{x})$ . We need two more things: $x^{2} + \frac{1}{x ^{2}}$ and $x + \frac{1}{x}$ .
2 $x^{2} + \frac{1}{x ^{2}} = 4^{2} + 2 = 18$ .
3 $(x + \frac{1}{x})^{2} = (x - \frac{1}{x})^{2} + 4 = 20$ , and $x > 0$ , so $x + \frac{1}{x} = 25$ .
4 $x^{4} - \frac{1}{x ^{4}} = 18 \times 25 \times 4 = 1445$ .

Answer 144√5

Check: the square of

x^{4} - \frac{1}{x ^{4}}

(x^{4} + \frac{1}{x ^{4}})^{2} - 4 = 32 2^{2} - 4 = 103680

and

(1445)^{2} = 103680

✓.

Practice

BasicTry it yourself first

x + \frac{1}{x} = 8

and

x > 1

, then

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

equals:

BasicTry it yourself first

x - \frac{1}{x} = 6

, then

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

equals:

StandardTry it yourself first

x - \frac{1}{x} = 5

, then

x^{5} - \frac{1}{x ^{5}}

equals:

StandardTry it yourself first

x - \frac{1}{x} = 4

, then

x^{6} + \frac{1}{x ^{6}}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 6

and

x > 1

, then

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

equals:

StandardTry it yourself first

x - \frac{1}{x} = 2

, then

x^{7} - \frac{1}{x ^{7}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = \frac{5}{2}

and

x > 1

, then

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

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 7

and

x > 1

, then

x^{4} - \frac{1}{x ^{4}}

equals:

AdvancedTry it yourself first

x - \frac{1}{x} = 7

, then

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

equals:

AdvancedTry it yourself first

x - \frac{1}{x} = 22

, then

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

equals:

Reaching the right form

Equation · fraction · coefficients · shifted variable · roots · x itself

How to bring a question into the form $x + \frac{1}{x}$ when it does not hand it to you directly.

You can climb the ladder only when $x + \frac{1}{x}$ (or $x - \frac{1}{x}$ ) is in your hand. The job of these six methods is to bring a question to that form: divide an equation by $x$ , flip and split a fraction, handle a pair with coefficients, rename a shifted variable, square a root form, and, if needed, find $x$ itself.

Dividing the equation by x

x^{2} - k x + 1 = 0 \Rightarrow x + \frac{1}{x} = k

When a quadratic (or quartic) equation is given instead of $x + \frac{1}{x}$ .

The idea

Say the equation $x^{2} - 7 x + 1 = 0$ is given. Its first and last coefficients match ( $1$ and $1$ ). $x = 0$ is not a solution (putting it in gives $1 \neq = 0$ ), so we may divide every term by $x$ : $x - 7 + \frac{1}{x} = 0$ . Move the $- 7$ across and $x + \frac{1}{x} = 7$ is ready. That is the start of the ladder, and we never needed $x$ itself.

◭ Diagram

One example (

k = 8

): dividing the quadratic by

x

makes

x + \frac{1}{x}

appear at once.

∑ Formula

a x^{2} + b x + a = 0 \Rightarrow x + \frac{1}{x} = - \frac{b}{a}

First = last coefficient

$a$: the coefficient of $x^{2}$ , equal to the constant term
$b$: the coefficient of $x$

∑ Formula

a x^{2} + b x - a = 0 \Rightarrow x - \frac{1}{x} = - \frac{b}{a}

Last coefficient is minus the first

$a$: the coefficient of $x^{2}$
$b$: the coefficient of $x$

∑ Formula

x^{4} + p x^{3} + q x^{2} + p x + 1 = 0 \div x^{2} a^{2} + p a + (q - 2) = 0, a = x + \frac{1}{x}

A quartic with matching end-to-end coefficients works the same way

$p, q$: the coefficients of the quartic
$a$: $x + \frac{1}{x}$ , which turns it into a quadratic

Equation	After dividing by $x$	Pair
$x^{2} - 9 x + 1 = 0$	$x - 9 + \frac{1}{x} = 0$	$x + \frac{1}{x} = 9$
$x^{2} + 1 = 6 x$	$x + \frac{1}{x} = 6$	$x + \frac{1}{x} = 6$
$2 x^{2} - 5 x + 2 = 0$	$2 x - 5 + \frac{2}{x} = 0$	$x + \frac{1}{x} = \frac{5}{2}$
$x^{2} + 6 x + 1 = 0$	$x + 6 + \frac{1}{x} = 0$	$x + \frac{1}{x} = - 6$
$x^{2} - 4 x - 1 = 0$	$x - 4 - \frac{1}{x} = 0$	$x - \frac{1}{x} = 4$
$3 x^{2} + 10 x - 3 = 0$	$3 x + 10 - \frac{3}{x} = 0$	$x - \frac{1}{x} = - \frac{10}{3}$

Here: the first column is the equation, the second its form after dividing by $x$ , the third the pair whose value we now hold. The last two rows show that a constant term equal to minus the first coefficient gives the minus pair.

Spot itSpot it: the first and last coefficients are equal (or equal with opposite signs) and the asked term is

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

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

…. Trap: if the coefficients are not equal (

x^{2} - 5 x + 2 = 0

x + \frac{1}{x}

does not appear directly; then you need another method (a pair with coefficients) or you must find

x

itself.

Steps

Check that the first and last coefficients match (or match with opposite signs).
Divide every term by $x$ (by $x^{2}$ for a quartic).
Put the $x$ and $\frac{1}{x}$ terms in a bracket and move the number across.
Now climb the ladder (square, cube, multiply).

Example 21 · Basic

x^{2} - 11 x + 1 = 0

, find

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

Solution, step by step

1The first and last coefficients are both $1$ . $x = 0$ is not a solution, so divide every term by $x$ : $x - 11 + \frac{1}{x} = 0$ .
2Move the $11$ across: $x + \frac{1}{x} = 11$ .
3Square: $x^{2} + \frac{1}{x ^{2}} = 1 1^{2} - 2 = 121 - 2 = 119$ .

Answer 119

The two values of

x

(

\frac{11 \pm 117}{2}

) are ugly; that is exactly why working without finding

x

is the way.

Example 22 · Standard

x^{2} + 1 = 9 x

, find

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

Solution, step by step

1Bring all terms to one side: $x^{2} - 9 x + 1 = 0$ ; dividing by $x$ gives $x + \frac{1}{x} = 9$ .
2 $x^{2} + \frac{1}{x ^{2}} = 9^{2} - 2 = 79$ .
3 $x^{4} + \frac{1}{x ^{4}} = 7 9^{2} - 2 = 6241 - 2 = 6239$ .

Answer 6239

The same ladder: for

x^{4}

take

x^{2}

as the base and square again.

Example 23 · Standard

2 x^{2} - 5 x + 2 = 0

and

x > 1

, find

x - \frac{1}{x}

Solution, step by step

1First and last coefficients $2$ and $2$ . Divide by $x$ : $2 x - 5 + \frac{2}{x} = 0$ , i.e. $2 (x + \frac{1}{x}) = 5$ , so $x + \frac{1}{x} = \frac{5}{2}$ .
2The bridge: $(x - \frac{1}{x})^{2} = \frac{25}{4} - 4 = \frac{9}{4}$ .
3As $x > 1$ , $x - \frac{1}{x} = \frac{3}{2}$ .

Answer 3/2

Check: the equation is

(2 x - 1) (x - 2) = 0

x = 2

and

2 - \frac{1}{2} = \frac{3}{2}

✓.

Example 24 · Advanced

x^{2} - 4 x - 1 = 0

, find

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

Solution, step by step

1The constant term is $- 1$ , the opposite of the first coefficient. Dividing by $x$ gives $x - 4 - \frac{1}{x} = 0$ , so $x - \frac{1}{x} = 4$ (a minus pair).
2The minus cube: $x^{3} - \frac{1}{x ^{3}} = 4^{3} + 3 \times 4 = 64 + 12 = 76$ .

Answer 76

Trap: do not write

x + \frac{1}{x} = 4

here; look at the sign of

- \frac{1}{x}

Example 25 · Advanced

x^{4} - 5 x^{3} + 8 x^{2} - 5 x + 1 = 0

and

x \neq = 1

, find

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

Solution, step by step

1The coefficients $1, - 5, 8, - 5, 1$ read the same from both ends. Divide by $x^{2}$ : $x^{2} - 5 x + 8 - \frac{5}{x} + \frac{1}{x ^{2}} = 0$ .
2Make pairs: $(x^{2} + \frac{1}{x ^{2}}) - 5 (x + \frac{1}{x}) + 8 = 0$ . Let $a = x + \frac{1}{x}$ ; then $(a^{2} - 2) - 5 a + 8 = 0$ , i.e. $a^{2} - 5 a + 6 = 0$ .
3 $(a - 2) (a - 3) = 0$ . $a = 2$ gives $x = 1$ , which is excluded. So $a = 3$ .
4 $x^{2} + \frac{1}{x ^{2}} = 3^{2} - 2 = 7$ .

Answer 7

Put

x = 1

in:

1 - 5 + 8 - 5 + 1 = 0

, so

x = 1

is a root, which explains why it was excluded.

Practice

BasicTry it yourself first

x^{2} - 12 x + 1 = 0

, then

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

equals:

BasicTry it yourself first

x^{2} + 6 x + 1 = 0

, then

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

equals:

StandardTry it yourself first

5 x^{2} - 26 x + 5 = 0

, then

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

equals:

StandardTry it yourself first

x^{2} - 6 x - 1 = 0

, then

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

equals:

StandardTry it yourself first

2 x^{2} + 5 x + 2 = 0

, then

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

equals:

StandardTry it yourself first

3 x^{2} - 8 x - 3 = 0

, then

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

equals:

AdvancedTry it yourself first

x^{4} + 1 = 7 x^{2}

and

x > 0

, then

x + \frac{1}{x}

equals:

AdvancedTry it yourself first

x^{4} - 7 x^{3} + 12 x^{2} - 7 x + 1 = 0

and

x \neq = 1

, then

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

equals:

AdvancedTry it yourself first

x^{2} - 25 x + 1 = 0

, then

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

equals:

AdvancedTry it yourself first

x^{2} + 10 x + 1 = 0

, then

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

equals:

Splitting fractions by dividing through

\frac{x}{x ^{2} + b x + 1} = \frac{1}{a + b}

When a fraction with $x^{2} + b x + 1$ or $x^{4} + k x^{2} + 1$ is given or asked.

The idea

Dividing numerator and denominator of a fraction by the same number does not change its value. Divide $x^{2} + b x + 1$ by $x$ and you get $x + b + \frac{1}{x}$ : the pair $x + \frac{1}{x}$ shows up by itself. So in $\frac{x}{x ^{2} + b x + 1}$ , dividing top and bottom by $x$ gives $\frac{1}{x + \frac{1}{x} + b}$ . In the same way $x^{4} + k x^{2} + 1$ divided by $x^{2}$ gives $x^{2} + \frac{1}{x ^{2}} + k$ .

∑ Formula

\frac{x ^{2} + b x + 1}{x} = a + b, \frac{x}{x ^{2} + b x + 1} = \frac{1}{a + b}

Divide by

x

$a$: the value of $x + \frac{1}{x}$
$b$: the middle coefficient (with its sign)

∑ Formula

\frac{x ^{4} + k x ^{2} + 1}{x ^{2}} = a^{2} - 2 + k, \frac{x ^{2}}{x ^{4} + k x ^{2} + 1} = \frac{1}{a ^{2} - 2 + k}

Divide by

x^{2}

$k$: the coefficient of $x^{2}$ (with its sign)
$a^{2} - 2$: $x^{2} + \frac{1}{x ^{2}}$

Fraction	After dividing through	Value
$\frac{x ^{2} + 1}{x}$	$x + \frac{1}{x}$	$a$
$\frac{x ^{4} + 1}{x ^{2}}$	$x^{2} + \frac{1}{x ^{2}}$	$a^{2} - 2$
$\frac{x ^{4} + x ^{2} + 1}{x ^{2}}$	$x^{2} + 1 + \frac{1}{x ^{2}}$	$a^{2} - 1$
$\frac{x ^{4} - x ^{2} + 1}{x ^{2}}$	$x^{2} - 1 + \frac{1}{x ^{2}}$	$a^{2} - 3$
$\frac{x ^{2}}{x ^{4} + x ^{2} + 1}$	$\frac{1}{x ^{2} + 1 + \frac{1}{x ^{2}}}$	$\frac{1}{a ^{2} - 1}$
$\frac{x}{x ^{2} - 3 x + 1}$	$\frac{1}{x - 3 + \frac{1}{x}}$	$\frac{1}{a - 3}$

Here: $a = x + \frac{1}{x}$ . The third column is the value in terms of $a$ . Note that $x^{4} + x^{2} + 1$ and $x^{4} - x^{2} + 1$ divided by $x^{2}$ give $a^{2} - 1$ and $a^{2} - 3$ , not $a^{2}$ .

ReverseThe reverse also appears: if

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

, flip the fraction:

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

, i.e.

x - 5 + \frac{1}{x} = 3

and

x + \frac{1}{x} = 8

Steps

Look at the highest power of $x$ in the fraction: if it is $x^{2}$ (like $x^{2} + b x + 1$ ) divide top and bottom by $x$ ; if it is $x^{4}$ (like $x^{4} + k x^{2} + 1$ ) divide by $x^{2}$ .
Divide every term of the top and bottom by that number; the $1$ becomes $\frac{1}{x}$ or $\frac{1}{x ^{2}}$ .
Form the pairs and put in $a$ or $a^{2} - 2$ ; for a reverse question flip the fraction first.

Example 26 · Basic

x + \frac{1}{x} = 7

, find

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

Solution, step by step

1Divide each term of the numerator by the denominator $x$ : $\frac{x ^{2}}{x} + \frac{3 x}{x} + \frac{1}{x} = x + 3 + \frac{1}{x}$ .
2Keep $x$ and $\frac{1}{x}$ together: $(x + \frac{1}{x}) + 3 = 7 + 3 = 10$ .

Answer 10

In the middle term

3 x

the

x

cancels and only

3

remains.

Example 27 · Standard

x + \frac{1}{x} = 9

, find

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

Solution, step by step

1The denominator is $x^{2} - 4 x + 1$ ; divide top and bottom by $x$ : $\frac{1}{x - 4 + \frac{1}{x}}$ .
2Denominator $= (x + \frac{1}{x}) - 4 = 9 - 4 = 5$ .
3So the value is $\frac{1}{5}$ .

Answer 1/5

The fraction is the reciprocal of the bracket; remember to take it.

Example 28 · Standard

x + \frac{1}{x} = 6

, find

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

Solution, step by step

1Divide numerator and denominator by $x^{2}$ : $\frac{1}{x ^{2} + 1 + \frac{1}{x ^{2}}}$ .
2 $x^{2} + \frac{1}{x ^{2}} = 6^{2} - 2 = 34$ , so the denominator is $34 + 1 = 35$ .
3The value is $\frac{1}{35}$ .

Answer 1/35

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

; this factorisation also leads to the same

a^{2} - 1

Example 29 · Advanced

\frac{x}{x ^{2} - 7 x + 1} = \frac{1}{4}

, find

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

Solution, step by step

1Flip both fractions: $\frac{x ^{2} - 7 x + 1}{x} = 4$ .
2Divide each term on the left by $x$ : $x - 7 + \frac{1}{x} = 4$ , so $x + \frac{1}{x} = 11$ .
3 $x^{2} + \frac{1}{x ^{2}} = 1 1^{2} - 2 = 119$ .

Answer 119

Flipping a fraction is fine only when neither side is zero.

Example 30 · Advanced

\frac{x ^{4} + x ^{2} + 1}{x ^{2}} = 48

and

x > 0

, find

x + \frac{1}{x}

Solution, step by step

1Dividing by $x^{2}$ : $x^{2} + 1 + \frac{1}{x ^{2}} = 48$ , so $x^{2} + \frac{1}{x ^{2}} = 47$ .
2 $(x + \frac{1}{x})^{2} = 47 + 2 = 49$ and $x > 0$ , so $x + \frac{1}{x} = 7$ .

Answer 7

You can also take

a^{2} - 1 = 48

straight to

a^{2} = 49

Practice

BasicTry it yourself first

x + \frac{1}{x} = 11

, then

\frac{x ^{2} + 5 x + 1}{x}

equals:

BasicTry it yourself first

x + \frac{1}{x} = 13

, then

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

equals:

StandardTry it yourself first

x + \frac{1}{x} = 8

, then

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

equals:

StandardTry it yourself first

x + \frac{1}{x} = 10

, then

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

equals:

StandardTry it yourself first

\frac{x}{x ^{2} - 9 x + 1} = \frac{1}{2}

, then

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

equals:

StandardTry it yourself first

x + \frac{1}{x} = 7

, then

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

equals:

AdvancedTry it yourself first

\frac{x ^{2}}{x ^{4} - x ^{2} + 1} = \frac{1}{13}

and

x > 0

, then

x + \frac{1}{x}

equals:

AdvancedTry it yourself first

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

, then

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

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 7

, then

\frac{x ^{4} + 3 x ^{2} + 1}{x ^{2}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = \frac{5}{2}

, then

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

equals:

Pairs with coefficients

u^{2} + w^{2} = S^{2} - 2 u w

When the pair looks like $3 x + \frac{2}{x}$ or $\frac{x}{3} + \frac{3}{x}$ and the product is not $1$ .

The idea

So far the pair was $x$ and $\frac{1}{x}$ , whose product is $1$ , so the square had a middle term $2$ . If the pair is $u$ and $w$ and their product is some number $c$ , the same reasoning works; just “ $2$ ” becomes “ $2 c$ ” and “ $3$ ” becomes “ $3 c$ ”: $(u + w)^{2} = u^{2} + w^{2} + 2 c$ and $(u + w)^{3} = u^{3} + w^{3} + 3 c (u + w)$ . The most important job is to find the product $u w$ . In $3 x + \frac{2}{x}$ , $u w = 3 x \cdot \frac{2}{x} = 6$ .

∑ Formula

u + w = S, u w = c \Rightarrow u^{2} + w^{2} = S^{2} - 2 c, u^{3} + w^{3} = S^{3} - 3 c S

The plus form

$u, w$: the two terms of the pair (like $3 x$ and $\frac{2}{x}$ )
$S$: their sum
$c$: their product, a number

∑ Formula

u - w = D, u w = c \Rightarrow u^{2} + w^{2} = D^{2} + 2 c, u^{3} - w^{3} = D^{3} + 3 cD

The minus form

$D$: their difference

Three situations arise. (a) The product is $1$ ( $\frac{x}{3} + \frac{3}{x}$ ): let $y = \frac{x}{3}$ and use the old ladder. (b) The product is $c \neq = 1$ and the asked pair is the same: use the formulas above. (c) The asked pair is different (given $3 a + \frac{1}{5 a}$ , asked $25 a^{2} + \frac{1}{9 a ^{2}}$ ): multiply the whole equation by a number so that the given terms become exactly the terms of the asked pair ( $5 a + \frac{1}{3 a}$ ), then do (b).

Spot itSpot it: the pair has coefficients. Multiply the two terms: if you get

1

it is (a), any other number means (b) or (c). Trap: do not feed such a pair straight into the

x + \frac{1}{x}

formulas.

Steps

Read the asked expression as $u^{2} + w^{2}$ (or $u^{3} \pm w^{3}$ ) and fix $u$ and $w$ .
Find the product $c = u w$ .
Multiply the given equation to get the value of $u \pm w$ .
Put $S$ and $c$ into the formula.

Example 31 · Basic

5 x + \frac{2}{x} = 9

, find

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

Solution, step by step

1 $25 x^{2} = (5 x)^{2}$ and $\frac{4}{x ^{2}} = (\frac{2}{x})^{2}$ , so $u = 5 x$ , $w = \frac{2}{x}$ .
2The product $u w = 5 x \times \frac{2}{x} = 10$ .
3Square: $(u + w)^{2} = u^{2} + w^{2} + 2 u w$ , so $u^{2} + w^{2} = 9^{2} - 2 \times 10 = 81 - 20 = 61$ .

Answer 61

10

in place of

1

: the middle term is now

20

, not

2

Example 32 · Standard

3 x - \frac{4}{x} = 8

, find

9 x^{2} + \frac{16}{x ^{2}}

Solution, step by step

1 $u = 3 x$ , $w = \frac{4}{x}$ , and the pair is a difference: $u - w = 8$ . Product $u w = 12$ .
2 $(u - w)^{2} = u^{2} + w^{2} - 2 u w$ , so $u^{2} + w^{2} = (u - w)^{2} + 2 u w = 64 + 24 = 88$ .

Answer 88

The square of a difference has

- 2 c

, so it becomes

+ 2 c

on the other side.

Example 33 · Standard

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

, find

\frac{x ^{2}}{9} + \frac{9}{x ^{2}}

and

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

Solution, step by step

1The product is $\frac{x}{3} \times \frac{3}{x} = 1$ , so take $y = \frac{x}{3}$ : $y + \frac{1}{y} = 7$ .
2 $\frac{x ^{2}}{9} + \frac{9}{x ^{2}} = y^{2} + \frac{1}{y ^{2}} = 7^{2} - 2 = 47$ .
3 $\frac{x ^{3}}{27} + \frac{27}{x ^{3}} = y^{3} + \frac{1}{y ^{3}} = 7^{3} - 3 \times 7 = 343 - 21 = 322$ .

Answer 47 ; 322

Naming

y

brought everything into the old formulas.

Example 34 · Advanced

3 a + \frac{1}{5 a} = 8

, find

25 a^{2} + \frac{1}{9 a ^{2}}

Solution, step by step

1The asked expression is $u^{2} + w^{2}$ with $u = 5 a$ and $w = \frac{1}{3 a}$ . But the given pair is $3 a$ and $\frac{1}{5 a}$ , which is different.
2To turn $3 a$ into $5 a$ multiply by $\frac{5}{3}$ : $5 a + \frac{5}{3} \times \frac{1}{5 a} = 5 a + \frac{1}{3 a}$ ✓, and the right side becomes $8 \times \frac{5}{3} = \frac{40}{3}$ .
3Now $u + w = \frac{40}{3}$ and $u w = 5 a \times \frac{1}{3 a} = \frac{5}{3}$ .
4 $u^{2} + w^{2} = (\frac{40}{3})^{2} - 2 \times \frac{5}{3} = \frac{1600}{9} - \frac{30}{9} = \frac{1570}{9}$ .

Answer 1570/9

Both terms must be multiplied by the same number; the second term came out right on its own because the ratio is the same.

Example 35 · Advanced

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

, find

8 x^{3} + \frac{1}{27 x ^{3}}

Solution, step by step

1 $8 x^{3} = (2 x)^{3}$ and $\frac{1}{27 x ^{3}} = (\frac{1}{3 x})^{3}$ . So the asked pair is the same: $u = 2 x$ , $w = \frac{1}{3 x}$ .
2 $u w = \frac{2}{3}$ and $u + w = 5$ .
3 $u^{3} + w^{3} = (u + w)^{3} - 3 u w (u + w) = 125 - 3 \times \frac{2}{3} \times 5 = 125 - 10 = 115$ .

Answer 115

3 c (u + w)

we get

3 \times \frac{2}{3} = 2

: the fraction cancelled.

Practice

BasicTry it yourself first

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

, then

16 x^{2} + \frac{9}{x ^{2}}

equals:

BasicTry it yourself first

2 x - \frac{5}{x} = 3

, then

4 x^{2} + \frac{25}{x ^{2}}

equals:

StandardTry it yourself first

\frac{x}{5} + \frac{5}{x} = 6

, then

\frac{x ^{2}}{25} + \frac{25}{x ^{2}}

equals:

StandardTry it yourself first

3 x + \frac{2}{x} = 7

, then

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

equals:

StandardTry it yourself first

a + \frac{1}{6 a} = 5

, then

36 a^{2} + \frac{1}{a ^{2}}

equals:

StandardTry it yourself first

3 x - \frac{2}{x} = 4

, then

27 x^{3} - \frac{8}{x ^{3}}

equals:

AdvancedTry it yourself first

2 x + \frac{3}{x} = 5

, then

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

equals:

AdvancedTry it yourself first

3 a + \frac{1}{4 a} = 8

, then

16 a^{2} + \frac{1}{9 a ^{2}}

equals:

AdvancedTry it yourself first

\frac{x}{2} + \frac{2}{x} = 9

, then

\frac{x ^{3}}{8} + \frac{8}{x ^{3}}

equals:

AdvancedTry it yourself first

7 a - \frac{1}{a} = 8

, then

343 a^{3} - \frac{1}{a ^{3}}

equals:

A shifted variable

Give

m = x - a

a new name

When the pair is built from $x \pm k$ , not from $x$ itself.

The idea

In many questions the “ $\frac{1}{something}$ ” pair is built not from $x$ but from a shifted term such as $(x - 2)$ or $(m + 3)$ . Give that term a new letter ( $n = x - 2$ ), write every other $x$ in terms of $n$ as well ( $x = n + 2$ ), and the question falls back into the old form: $n + \frac{1}{n}$ or $n - \frac{1}{n}$ . The new letter is the one under the “ $1$ ” or the one repeated in the asked expression.

∑ Formula

m + \frac{1}{m - k} = S \Rightarrow n + \frac{1}{n} = S - k (n = m - k)

The term under the fraction is the new variable

$m$: the given variable
$k$: the shift (how much was subtracted)
$S$: the given sum
$n$: the new variable $m - k$

∑ Formula

(x - a) (x - b) = 1 \Rightarrow m - \frac{1}{m} = b - a (m = x - a)

Two factors with product

1

$a, b$: the constants in the two factors
$m$: the new variable $x - a$

Third form: a quadratic $x^{2} + p x = r$ is given and the question asks $(x + k)^{2} + \frac{1}{( x + k ) ^{2}}$ . Put $m = x + k$ and the equation turns into $m^{2} - S m + 1 = 0$ (this happens exactly when $S = 2 k - p$ and $r = k^{2} - p k - 1$ ). The hint is that the asked expression rests on $x + k$ .

Special caseWhen

n + \frac{1}{n} = 2

comes out,

n = 1

(and

n = - 1

for

- 2

). Then every power of that term is

1

and all questions about

n

open without the ladder.

Steps

Decide which term becomes the new letter (the one under $1$ or in the asked expression).
Write $x$ (or $m$ ) in terms of the new letter and substitute it into the equation.
Simplify and divide by the new letter: $n \pm \frac{1}{n} =$ a number.
Climb the ladder; if the asked expression is in $x$ , convert back at the end.

Example 36 · Basic

m + \frac{1}{m - 3} = 8

, find

(m - 3)^{2} + \frac{1}{( m - 3 ) ^{2}}

Solution, step by step

1The denominator is $m - 3$ , so take $n = m - 3$ ; then $m = n + 3$ .
2Put it in: $n + 3 + \frac{1}{n} = 8$ , i.e. $n + \frac{1}{n} = 5$ .
3Square: $n^{2} + \frac{1}{n ^{2}} = 5^{2} - 2 = 23$ .

Answer 23

The asked expression rests on the same

(m - 3)

, so the answer came straight in

n

; there is no need to find

m

Example 37 · Standard

x^{2} + x = 11

, find

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

Solution, step by step

1The asked expression rests on $x + 4$ , so $m = x + 4$ , i.e. $x = m - 4$ .
2Put it into the equation: $(m - 4)^{2} + (m - 4) = 11$ , i.e. $m^{2} - 8 m + 16 + m - 4 = 11$ .
3Simplifying gives $m^{2} - 7 m + 1 = 0$ (since $16 - 4 - 11 = 1$ ). Divide by $m$ : $m + \frac{1}{m} = 7$ .
4 $m^{2} + \frac{1}{m ^{2}} = 7^{2} - 2 = 47$ .

Answer 47

If the constant had not come out as

1

, this method would not produce

m + \frac{1}{m}

; that is the question-setter’s design.

Example 38 · Standard

(x - 2) (x - 5) = 1

, find

(x - 2)^{3} - \frac{1}{( x - 2 ) ^{3}}

Solution, step by step

1Take $m = x - 2$ . Then $x - 5 = m - 3$ .
2The equation: $m (m - 3) = 1$ . Divide by $m$ : $m - 3 = \frac{1}{m}$ , so $m - \frac{1}{m} = 3$ .
3The minus cube: $m^{3} - \frac{1}{m ^{3}} = 3^{3} + 3 \times 3 = 27 + 9 = 36$ .

Answer 36

Rule:

(x - a) (x - b) = 1 \Rightarrow m - \frac{1}{m} = b - a

; here

5 - 2 = 3

Example 39 · Advanced

m + \frac{1}{m - 4} = 6

, find

m^{3} - m

and

(m - 4)^{7} + \frac{1}{( m - 4 ) ^{7}}

Solution, step by step

1Take $n = m - 4$ , $m = n + 4$ : $n + 4 + \frac{1}{n} = 6$ , so $n + \frac{1}{n} = 2$ .
2 $n + \frac{1}{n} = 2$ means $(n - 1)^{2} = 0$ , so $n = 1$ and $m = 5$ .
3 $m^{3} - m = 125 - 5 = 120$ .
4Every power of $n = 1$ is $1$ , so $(m - 4)^{7} + \frac{1}{( m - 4 ) ^{7}} = 1 + 1 = 2$ (no ladder needed).

Answer 120 ; 2

Here

m

itself came out, so any question about

m

is solved by putting in the number.

Practice

BasicTry it yourself first

m + \frac{1}{m - 2} = 9

, then

(m - 2)^{2} + \frac{1}{( m - 2 ) ^{2}}

equals:

BasicTry it yourself first

x^{2} - x = 5

, then

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

equals:

StandardTry it yourself first

(x - 3) (x - 7) = 1

, then

(x - 3)^{2} + \frac{1}{( x - 3 ) ^{2}}

equals:

StandardTry it yourself first

m + \frac{1}{m + 3} = 1

, then

(m + 3)^{3} + \frac{1}{( m + 3 ) ^{3}}

equals:

StandardTry it yourself first

x^{2} - 4 x = 4

, then

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

equals:

StandardTry it yourself first

(x - 1) (x - 7) = 1

, then

(x - 1)^{3} - \frac{1}{( x - 1 ) ^{3}}

equals:

AdvancedTry it yourself first

m + \frac{1}{m - 6} = 8

, then

m^{2} + m - 1

equals:

AdvancedTry it yourself first

x^{2} + 2 x = 14

, then

(x + 5)^{4} + \frac{1}{( x + 5 ) ^{4}}

equals:

AdvancedTry it yourself first

m + \frac{1}{m - 1} = 7

, then

(m - 1)^{3} + \frac{1}{( m - 1 ) ^{3}}

equals:

AdvancedTry it yourself first

m + \frac{1}{m - 2} = 4

, then

m^{3} - m

equals:

Square-root and cube-root forms

Square of

x + \frac{1}{x}

When the pair is built from $x$ or $3 x$ , or when its value is asked.

The idea

$x$ and $\frac{1}{x}$ also multiply to $1$ , so this is the same pair with “ $x$ ” in place of “ $x$ ”. Squaring turns $x$ into $x$ : $(x + \frac{1}{x})^{2} = x + 2 + \frac{1}{x}$ . So the pair one rung below gives the value one rung above. With a cube root, the cube formula runs in reverse in the same way.

∑ Formula

x + \frac{1}{x} = σ \Rightarrow x + \frac{1}{x} = σ^{2} - 2, x - \frac{1}{x} = τ \Rightarrow x + \frac{1}{x} = τ^{2} + 2

The square-root pair

$σ, τ$: the given values: sum and difference of the $x$ pair
$x + \frac{1}{x}$: the value one rung above

∑ Formula

3 x + \frac{1}{3 x} = σ \Rightarrow x + \frac{1}{x} = σ^{3} - 3 σ, 3 x - \frac{1}{3 x} = τ \Rightarrow x - \frac{1}{x} = τ^{3} + 3 τ

The cube-root pair

$σ, τ$: sum and difference of the cube-root pair

∑ Formula

x + \frac{1}{x} = s (x > 0) \Rightarrow x + \frac{1}{x} = s + 2, x - \frac{1}{x} = s - 2 (x > 1)

The reverse direction

$s$: the value of $x + \frac{1}{x}$
$x > 0, x > 1$: conditions that settle the sign

Given	You get
$x + \frac{1}{x} = σ$	$x + \frac{1}{x} = σ^{2} - 2$
$x - \frac{1}{x} = τ$	$x + \frac{1}{x} = τ^{2} + 2$
$3 x + \frac{1}{3 x} = σ$	$x + \frac{1}{x} = σ^{3} - 3 σ$
$3 x - \frac{1}{3 x} = τ$	$x - \frac{1}{x} = τ^{3} + 3 τ$
$x + \frac{1}{x} = s$	$x^{3/2} + \frac{1}{x ^{3/2}} = σ^{3} - 3 σ$ with $σ = s + 2$

Here: in the first four rows the given pair is on the left and the value of the next bigger power on the right. In the last row $x^{3/2}$ is asked: first find $σ$ (because $x^{3/2} = (x)^{3}$ ), then the cube formula.

Spot itSpot it: the question shows

x

x^{1/3}

x^{3/2}

and the pair is tied to

1

. Read the sign condition: without

x > 0

the

x

is not real.

Steps

Identify the power the pair is built on ( $x$ , $3 x$ or $x$ ).
For a square root, square (subtract $2$ for a sum, add $2$ for a difference); for a cube root, use the cube formula.
Now $x + \frac{1}{x}$ is in hand: climb the ladder or build the asked expression.

Example 40 · Basic

x + \frac{1}{x} = 7

, find

x + \frac{1}{x}

and

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

Solution, step by step

1Square: $(x + \frac{1}{x})^{2} = x + 2 + \frac{1}{x} = 49$ , so $x + \frac{1}{x} = 47$ .
2One more square: $x^{2} + \frac{1}{x ^{2}} = 4 7^{2} - 2 = 2209 - 2 = 2207$ .

Answer 47 ; 2207

Each squaring doubles the power:

x^{1/2} \to x \to x^{2}

Example 41 · Standard

3 x + \frac{1}{3 x} = 4

, find

x + \frac{1}{x}

Solution, step by step

1Let $y = 3 x$ . Then $y + \frac{1}{y} = 4$ and $x = y^{3}$ , so $x + \frac{1}{x} = y^{3} + \frac{1}{y ^{3}}$ .
2The cube formula: $y^{3} + \frac{1}{y ^{3}} = 4^{3} - 3 \times 4 = 64 - 12 = 52$ .

Answer 52

Here the given pair is cubed, not squared.

Example 42 · Standard

x + \frac{1}{x} = 18

and

x > 1

, find

x + \frac{1}{x}

and

x - \frac{1}{x}

Solution, step by step

1The square of the sum: $(x + \frac{1}{x})^{2} = 18 + 2 = 20$ , and $x > 1$ ⟹ positive: $20 = 25$ .
2The square of the difference: $(x - \frac{1}{x})^{2} = 18 - 2 = 16$ , and $x > 1$ ⟹ positive: $4$ .

Answer 2√5 ; 4

Check with the bridge:

(25)^{2} - 4^{2} = 20 - 16 = 4

✓.

Example 43 · Advanced

x - \frac{1}{x} = 3

, find

x + \frac{1}{x}

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

and

x^{3/2} - \frac{1}{x ^{3/2}}

Solution, step by step

1Square: $(x - \frac{1}{x})^{2} = x - 2 + \frac{1}{x} = 9$ , so $x + \frac{1}{x} = 11$ .
2One more square: $x^{2} + \frac{1}{x ^{2}} = 1 1^{2} - 2 = 119$ .
3 $x^{3/2} = (x)^{3}$ , so the minus cube: $(x)^{3} - \frac{1}{( x ) ^{3}} = 3^{3} + 3 \times 3 = 27 + 9 = 36$ .

Answer 11 ; 119 ; 36

All three answers come from the one given value, on different rungs.

Practice

BasicTry it yourself first

x + \frac{1}{x} = 9

, then

x + \frac{1}{x}

equals:

BasicTry it yourself first

x - \frac{1}{x} = 4

, then

x + \frac{1}{x}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 14 (x > 0)

, then

x + \frac{1}{x}

equals:

StandardTry it yourself first

3 x + \frac{1}{3 x} = 5

, then

x + \frac{1}{x}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 23

, then

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

equals:

StandardTry it yourself first

x + \frac{1}{x} = 11 (x > 1)

, then

x - \frac{1}{x}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 23 (x > 0)

, then

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

equals:

AdvancedTry it yourself first

3 x - \frac{1}{3 x} = 2

, then

x - \frac{1}{x}

equals:

AdvancedTry it yourself first

x - \frac{1}{x} = 6

, then

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

equals:

Finding x itself and values like 2 + √3

The conjugate is the reciprocal

When $x$ is given directly as $p + q$ , or must be found.

The idea

Sometimes $x$ itself is given, like $x = 2 + 3$ , and $x + \frac{1}{x}$ or $x^{2} + \frac{1}{x ^{2}}$ is asked. Then we need $\frac{1}{x}$ , which comes from rationalising: $\frac{1}{2 + 3} = \frac{2 - 3}{( 2 + 3 ) ( 2 - 3 )} = \frac{2 - 3}{4 - 3} = 2 - 3$ . The denominator came out $1$ , so the conjugate is the reciprocal, and $x + \frac{1}{x} = 4$ , $x - \frac{1}{x} = 23$ at once. After that the ladder is the old one. In the reverse question ( $x$ from $x + \frac{1}{x} = s$ ) the quadratic formula helps: $x = \frac{s \pm s ^{2} - 4}{2}$ ; the two roots are reciprocals of each other.

∑ Formula

x = p + q, p^{2} - q = 1 \Rightarrow \frac{1}{x} = p - q, x + \frac{1}{x} = 2 p, x - \frac{1}{x} = 2 q

The product-

1

form

$p, q$: the numbers in the given $x$ ( $q$ is under the root)
$p^{2} - q$: the value of the denominator $(p + q) (p - q)$

∑ Formula

x = p + q, p^{2} - q = - 1 \Rightarrow \frac{1}{x} = q - p, x - \frac{1}{x} = 2 p, x + \frac{1}{x} = 2 q

The product-

(- 1)

form

$p, q$: the same numbers

∑ Formula

x = a + b, a - b = 1 \Rightarrow \frac{1}{x} = a - b, x + \frac{1}{x} = 2 a, x - \frac{1}{x} = 2 b

The two-roots form

$a, b$: the numbers under the roots, with difference $1$

$x$	Denominator	$\frac{1}{x}$	$x + \frac{1}{x}$
$2 + 3$	$4 - 3 = 1$	$2 - 3$	$4$
$3 + 22$	$9 - 8 = 1$	$3 - 22$	$6$
$5 + 26$	$25 - 24 = 1$	$5 - 26$	$10$
$2 + 5$	$4 - 5 = - 1$	$5 - 2$	$25$
$\frac{3 + 5}{2}$	$\frac{9 - 5}{4} = 1$	$\frac{3 - 5}{2}$	$3$

Here: the last column is the value of $x + \frac{1}{x}$ ; note that for a $- 1$ case like $2 + 5$ , $x - \frac{1}{x}$ is the whole number ( $4$ ) and $x + \frac{1}{x}$ the root ( $25$ ): the roles swap. If the denominator is neither $1$ nor $- 1$ , just write $\frac{1}{x}$ by direct rationalising.

Spot itSpot it: in

x = p + q

compute

p^{2} - q

; if it is

\pm 1

the conjugate is the reciprocal. For

7 + 43

write

43 = 48

and take

q = 48

49 - 48 = 1

. Tip: of

x + \frac{1}{x}

and

x - \frac{1}{x}

, start the ladder from whichever comes out a whole number.

Steps

Write $x$ as $p + q k$ (if needed, $q k = q^{2} k$ ).
Find the denominator $p^{2} - q^{2} k$ . If it is $1$ or $- 1$ , $\frac{1}{x}$ is the conjugate (or minus it); otherwise rationalise.
Write $x + \frac{1}{x}$ and $x - \frac{1}{x}$ ; then climb the ladder.

Example 44 · Basic

x = 5 + 26

, find

x + \frac{1}{x}

and

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

Solution, step by step

1For $\frac{1}{x}$ multiply by the conjugate $5 - 26$ : the denominator is $25 - 24 = 1$ , so $\frac{1}{x} = 5 - 26$ .
2Sum: $x + \frac{1}{x} = (5 + 26) + (5 - 26) = 10$ (the roots cancel).
3Square: $x^{2} + \frac{1}{x ^{2}} = 1 0^{2} - 2 = 98$ .

Answer 10 ; 98

Expanding the square directly also works, but the ladder keeps the numbers small.

Example 45 · Standard

x = 7 + 6

, find

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

and

x - \frac{1}{x}

Solution, step by step

1 $(7 + 6) (7 - 6) = 7 - 6 = 1$ , so $\frac{1}{x} = 7 - 6$ .
2 $x + \frac{1}{x} = 27$ and $x - \frac{1}{x} = 26$ .
3 $x^{2} + \frac{1}{x ^{2}} = (27)^{2} - 2 = 28 - 2 = 26$ .

Answer 26 ; 2√6

Check with the other pair:

(26)^{2} + 2 = 24 + 2 = 26

✓.

Example 46 · Standard

x = 3 + 22

, find

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

Solution, step by step

1The denominator is $9 - 8 = 1$ , so $\frac{1}{x} = 3 - 22$ .
2Difference: $x - \frac{1}{x} = 42$ (the $3$ s cancel and $22 + 22$ remains).
3The minus cube: $x^{3} - \frac{1}{x ^{3}} = (42)^{3} + 3 \times 42 = 1282 + 122 = 1402$ .

Answer 140√2

(42)^{3} = 64 \times 22 = 1282

Example 47 · Advanced

x = 4 + 17

, find

x - \frac{1}{x}

x + \frac{1}{x}

and

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

Solution, step by step

1Denominator: $(4 + 17) (4 - 17) = 16 - 17 = - 1$ . So $\frac{1}{x} = \frac{4 - 17}{- 1} = 17 - 4$ .
2Difference: $x - \frac{1}{x} = (4 + 17) - (17 - 4) = 8$ . Sum: $x + \frac{1}{x} = (4 + 17) + (17 - 4) = 217$ .
3 $x^{2} - \frac{1}{x ^{2}} = (x + \frac{1}{x}) (x - \frac{1}{x}) = 217 \times 8 = 1617$ .

Answer 8 ; 2√17 ; 16√17

Trap: it would be wrong to expect

x + \frac{1}{x}

to be the whole number here; with

- 1

the roles swap.

Example 48 · Advanced

x + \frac{1}{x} = 8

and

x > 1

, find

x

and

x - \frac{1}{x}

Solution, step by step

1Multiply by $x$ : $x^{2} - 8 x + 1 = 0$ . Quadratic formula: $x = \frac{8 \pm 64 - 4}{2} = \frac{8 \pm 60}{2} = 4 \pm 15$ .
2The roots multiply to $16 - 15 = 1$ , i.e. they are reciprocals. As $x > 1$ , $x = 4 + 15$ .
3 $\frac{1}{x} = 4 - 15$ , so $x - \frac{1}{x} = 215$ .

Answer 4+√15 ; 2√15

Also by the bridge:

64 - 4 = 60 = 215

✓.

Practice

BasicTry it yourself first

x = 9 + 45

, then

x + \frac{1}{x}

equals:

BasicTry it yourself first

x = 2 + 3

, then

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

equals:

StandardTry it yourself first

x = 3 + 10

, then

x - \frac{1}{x}

equals:

StandardTry it yourself first

x = \frac{3 + 5}{2}

, then

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

equals:

StandardTry it yourself first

x = 6 + 5

, then

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

equals:

StandardTry it yourself first

x + \frac{1}{x} = 12

and

x > 1

, then

x

equals:

AdvancedTry it yourself first

x = 11 + 230

, then

x + \frac{1}{x}

equals:

AdvancedTry it yourself first

x = \frac{1 + 5}{2}

, then

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

equals:

AdvancedTry it yourself first

x = 6 + 37

, then

x - \frac{1}{x}

equals:

AdvancedTry it yourself first

x = 2 + 3

, then

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

equals:

Summary, mixed practice, revision

All methods in one place

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

Here is the essence of the chapter and exam-style mixed questions, where you first spot the method and then solve.

Summary of all methods

Which method when

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

Method	Use when	Key formula / move
The square	$x^{2} + \frac{1}{x ^{2}}$ from $x \pm \frac{1}{x}$	$a^{2} - 2$ , $b^{2} + 2$ ; $(a)^{2} - (b)^{2} = 4$
Going back	the $x^{2}$ or $x^{4}$ value is given, $x \pm \frac{1}{x}$ wanted	$a + 2$ , $a - 2$ ; sign from the condition
The cube	$x^{3} \pm \frac{1}{x ^{3}}$	$a^{3} - 3 a$ , $b^{3} + 3 b$
Higher powers	$x^{4}, \dots, x^{8}$	$a_{m + n} = a_{m} a_{n} - a_{m - n}$
The minus family	$x^{n} - \frac{1}{x ^{n}}$	factorise squares/cubes; $u_{n + 1} = b u_{n} + u_{n - 1}$
Divide by $x$	an equation with matching coefficients	$a x^{2} + b x + a = 0 \Rightarrow x + \frac{1}{x} = - \frac{b}{a}$
Fractions	$\frac{x}{x ^{2} + b x + 1}$ , $\frac{x ^{2}}{x ^{4} + k x ^{2} + 1}$	$\frac{1}{a + b}$ , $\frac{1}{a ^{2} - 2 + k}$
Pairs with coefficients	$3 x + \frac{2}{x}$ , $\frac{x}{3} + \frac{3}{x}$	$u^{2} + w^{2} = S^{2} - 2 u w$
Shifted variable	terms like $(x - 2)$ , $\frac{1}{m - 3}$	name it $n$ ; make $n \pm \frac{1}{n}$
Root forms	$x$ , $3 x$ , $x^{3/2}$	$σ^{2} - 2$ , $σ^{3} - 3 σ$
$x$ itself	$x = p + q$ , or find $x$	denominator $p^{2} - q = \pm 1$ ⟹ $\frac{1}{x}$ = conjugate

Here: $a = x + \frac{1}{x}$ , $b = x - \frac{1}{x}$ , $a_{n} = x^{n} + \frac{1}{x ^{n}}$ , $u_{n} = x^{n} + (- \frac{1}{x})^{n}$ , $S = u + w$ and $σ$ is the value of a root pair. First ask: which pair is it, and is its product $1$ or some other number?

Two slipsThe two biggest slips: (1) using

- 2

with a minus pair (it takes

+ 2

); (2) taking a square root and forgetting the sign condition. Before answering, check each time:

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

and

x > 1

⟹

x - \frac{1}{x} > 0

Mixed practice — set 1

Basic to standard

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

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

BasicTry it yourself first

x + \frac{1}{x} = 12

, then

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

equals:

BasicTry it yourself first

x - \frac{1}{x} = 7

, then

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

equals:

BasicTry it yourself first

x + \frac{1}{x} = 12

, then

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

equals:

BasicTry it yourself first

x^{2} - 14 x + 1 = 0

, then

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

equals:

BasicTry it yourself first

x + \frac{1}{x} = 12

, then

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

equals:

BasicTry it yourself first

x + \frac{1}{x} = 8

, then

x + \frac{1}{x}

equals:

BasicTry it yourself first

x = 4 + 15

, then

x + \frac{1}{x}

equals:

StandardTry it yourself first

x^{2} + \frac{1}{x ^{2}} = 119

and

x > 0

, then

x + \frac{1}{x}

equals:

StandardTry it yourself first

3 x + \frac{4}{x} = 8

, then

9 x^{2} + \frac{16}{x ^{2}}

equals:

StandardTry it yourself first

m + \frac{1}{m - 3} = 10

, then

(m - 3)^{2} + \frac{1}{( m - 3 ) ^{2}}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 9

, then

x^{5} + \frac{1}{x ^{5}}

equals:

StandardTry it yourself first

x + \frac{1}{x} = 12

and

x > 1

, then

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

equals:

StandardTry it yourself first

x - \frac{1}{x} = 4

, then

x^{5} - \frac{1}{x ^{5}}

equals:

StandardTry it yourself first

x^{2} + x = 19

, then

(x + 5)^{2} + \frac{1}{( x + 5 ) ^{2}}

equals:

StandardTry it yourself first

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

, then

x + \frac{1}{x}

equals:

StandardTry it yourself first

(x - 4) (x - 9) = 1

, then

(x - 4)^{2} + \frac{1}{( x - 4 ) ^{2}}

equals:

StandardTry it yourself first

x^{4} + 1 = 14 x^{2}

and

x > 0

, then

x + \frac{1}{x}

equals:

StandardTry it yourself first

x^{2} + \frac{1}{x ^{2}} = 23

and

x > 1

, then

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

equals:

Mixed practice — set 2

Standard to advanced

Mixed questions from the hard end of the exam.

AdvancedTry it yourself first

x + \frac{1}{x} = 9

, then

x^{6} + \frac{1}{x ^{6}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 10

, then

x^{7} + \frac{1}{x ^{7}}

equals:

AdvancedTry it yourself first

x - \frac{1}{x} = 3

, then

x^{7} - \frac{1}{x ^{7}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 12

and

x > 1

, then

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

equals:

AdvancedTry it yourself first

x^{4} - 8 x^{3} + 14 x^{2} - 8 x + 1 = 0

and

x \neq = 1

, then

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

equals:

AdvancedTry it yourself first

3 x^{2} + 10 x + 3 = 0

, then

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

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 9

, then

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

equals:

AdvancedTry it yourself first

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

, then

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

equals:

AdvancedTry it yourself first

\frac{x ^{2}}{x ^{4} + 2 x ^{2} + 1} = \frac{1}{25}

and

x > 0

, then

x + \frac{1}{x}

equals:

AdvancedTry it yourself first

2 a + \frac{1}{3 a} = 7

, then

9 a^{2} + \frac{1}{4 a ^{2}}

equals:

AdvancedTry it yourself first

5 x - \frac{2}{x} = 11

, then

125 x^{3} - \frac{8}{x ^{3}}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 98 (x > 0)

, then

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

equals:

AdvancedTry it yourself first

x = 2 + 5

, then

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

equals:

AdvancedTry it yourself first

x = 5 + 26

, then

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

equals:

AdvancedTry it yourself first

x^{2} - 2 x = 47

, then

(x + 6)^{3} + \frac{1}{( x + 6 ) ^{3}}

equals:

AdvancedTry it yourself first

m + \frac{1}{m + 2} = 3

, then

(m + 2)^{4} + \frac{1}{( m + 2 ) ^{4}}

equals:

AdvancedTry it yourself first

x = 15 + 414

, then

x + \frac{1}{x}

equals:

AdvancedTry it yourself first

x^{4} + \frac{1}{x ^{4}} = 3842

and

x > 0

, then

x + \frac{1}{x}

equals:

AdvancedTry it yourself first

x + \frac{1}{x} = 9

, which of these is

x^{5} + \frac{1}{x ^{5}}

AdvancedTry it yourself first

x + \frac{1}{x} = 16

and

x > 1

, then

x

equals:

Revision: cards and traps

Flip and recall

Read the front, think, then flip.

NextNext step: the chapter “x + 1/x: Cycles, Limits and Several Variables” has special values such as

x + \frac{1}{x} = 1, - 1, 3

, huge powers (

x^{100}

), the consequences of

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

, minimum values and three-variable questions.