Algebra

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Subtopic 2 of 6·← Basic algebraic identities·$a^3+b^3+c^3-3abc$ and conditional identities →

$x+\frac{1}{x}$ type expressions

If $x+\frac1x=k$ is known, every expression of the form $x^n+\frac{1}{x^n}$ can be built from $k$ without solving for $x$ . The same works for $x-\frac1x$ and for mixed forms like $2x+\frac{1}{3x}$ .

How it is disguised in CGL:

a quadratic like $x^2-4x+1=0$ (divide by $x$ : $x+\frac1x=4$ );
a surd like $x=2+\sqrt3$ (then $\frac1x=2-\sqrt3$ );
a special value: $x+\frac1x=1,\ -1,\ \sqrt3,\ \sqrt2$ gives cyclic powers of $x$ .

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 14 practice Q

Detailed notes

The ladder

If $x+\dfrac1x = k$ , then every $x^n+\dfrac1{x^n}$ climbs from $k$ without ever finding $x$ : $x^2+\frac1{x^2} = k^2-2 \qquad x^3+\frac1{x^3} = k^3-3k \qquad x^4+\frac1{x^4} = (x^2+\tfrac1{x^2})^2-2$ and the multiply-down shortcut for the fifth rung: $x^5+\frac1{x^5} = \left(x^2+\frac1{x^2}\right)\left(x^3+\frac1{x^3}\right) - \left(x+\frac1x\right)$ With $k=3$ : $x^2+\frac1{x^2} = 7$ , $x^3+\frac1{x^3} = 18$ , $x^5+\frac1{x^5} = 7\times18-3 = 123$ . Squaring again gives the fourth rung: $7^2-2 = 47$ .

The difference ladder

For $x-\dfrac1x = m$ the minus version works the same way, with +2 in the square: $x^2+\frac1{x^2} = m^2+2 \qquad x^3-\frac1{x^3} = m^3+3m \qquad x^4+\frac1{x^4} = (m^2+2)^2-2$ And the two ladders connect: $\left(x+\frac1x\right)^2 = \left(x-\frac1x\right)^2+4$ . Given either one, the other is one square away — $x+\frac1x = \sqrt{m^2+4}$ .

Quadratics with equal end coefficients

$4x^2 - 9x + 4 = 0$ — first and last coefficient equal — divides cleanly by $x$ : $4x - 9 + \frac4x = 0 \;\Rightarrow\; x+\frac1x = \frac94$ So the quadratic IS the $k$ -value in disguise. After that, run the ladder. Any $ax^2+bx+a=0$ gives $x+\frac1x = \frac ba$ .

Special values — the cyclic escape

Some $k$ values collapse the ladder entirely because $x$ satisfies a cyclotomic quadratic:

$x+\frac1x = 1$ → $x^2-x+1 = 0$ → $x^3 = -1$ → $x^6 = x^9 = \dots = 1$ .
$x+\frac1x = -1$ → $x^2+x+1 = 0$ → $x^3 = 1$ .
$x+\frac1x = \sqrt3$ → $x^2+\frac1{x^2} = -1$ → $x^4 = -1$ -type cycling ( $x^6+\frac1{x^6} = -2$ , $x^{18}$ -type sums vanish to small integers). For stems like $x^{18}+x^{12}+x^6+1$ , reduce every exponent mod 3 (or mod 6) before doing anything.

Which rung to compute

Plan the route BEFORE multiplying anything. For $k=5$ asking $x^4+\frac1{x^4}$ : two squarings — $23$ , then $23^2-2 = 527$ — never touch a cube. For fifth rungs, multiply rung 2 by rung 3 and subtract $k$ ; do not compute the sixth rung by cubing rung two. When a stem mixes sum and difference ladders (given $x-\frac1x$ , asked about $x^3+\frac1{x^3}$ ), convert once with $k^2 = m^2+4$ , then stay on the ladder that matches the asked sign.

Common traps

Using $k^2-2$ for the difference ladder (it is $m^2+2$ ).
Writing $x^3+\frac1{x^3} = k^3-3$ instead of $k^3-3k$ .
Dividing a quadratic by $x$ when the end coefficients differ — that trick needs $ax^2+bx+a$ .
For a cyclic stem, expanding powers instead of reducing exponents first.

Quick revision

$k = x+\frac1x$ : $x^2+\frac1{x^2} = k^2-2$ , $x^3+\frac1{x^3} = k^3-3k$ , $x^5+\frac1{x^5} = (k^2-2)(k^3-3k)-k$ .
$m = x-\frac1x$ : $x^2+\frac1{x^2} = m^2+2$ , $x^3-\frac1{x^3} = m^3+3m$ ; $k^2 = m^2+4$ .
$ax^2+bx+a=0$ → $x+\frac1x = \frac ba$ .
$k = 1$ → $x^3 = -1$ ; $k = -1$ → $x^3 = 1$ ; $k = \sqrt3$ → $x^2+\frac1{x^2} = -1$ .

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Ladder from $x+\frac1x=k$very common4 practice Q

How to spot it:

$x+\frac1x = k$ given; $x^2+\frac1{x^2}$ , $x^3+\frac1{x^3}$ , $x^4+\frac1{x^4}$ or $x^5+\frac1{x^5}$ asked.

x^2+\frac1{x^2} = k^2-2; \quad x^3+\frac1{x^3} = k^3-3k; \quad x^4+\frac1{x^4} = (k^2-2)^2-2

Square once: $x^2+\frac1{x^2} = k^2-2$ (the $2x\cdot\frac1x = 2$ cancels).
Cube for the third rung: $k^3-3k$ .
Fourth rung: square the second. Fifth rung: multiply the second and third and subtract $k$ .

Why: the product $x\cdot\frac1x = 1$ makes every rung a polynomial in $k$ alone.

Example: If $x+\frac1x=6$ , then $x^3+\frac1{x^3}$ is:

$6^3 - 3\times6 = 216 - 18 = 198$ . One line, no solving for $x$ .

Type 2: Difference ladder from $x-\frac1x=m$very common2 practice Q

How to spot it:

$x-\frac1x = m$ given; the squared rung or $x^3-\frac1{x^3}$ is asked.

x^2+\frac1{x^2} = m^2+2; \quad x^3-\frac1{x^3} = m^3+3m; \quad k^2 = m^2+4

Square: the cross term is $-2$ but it SUBTRACTS, so $x^2+\frac1{x^2} = m^2+2$ .
Cube: $x^3-\frac1{x^3} = m^3+3m$ (the correction adds).
To switch ladders: $(x+\frac1x)^2 = (x-\frac1x)^2+4$ .

Why: same expansions as the sum ladder with the sign of the cross term flipped.

Example: If $x-\frac1x=4$ , then $x^3-\frac1{x^3}$ is:

$4^3 + 3\times4 = 64+12 = 76$ . (And $x^2+\frac1{x^2} = 18$ . $x+\frac1x = \sqrt{20} = 2\sqrt5$ .)

Type 3: Quadratic with equal end coefficientsvery common2 practice Q

How to spot it:

An equation like $4x^2-9x+4=0$ (equal first and last coefficients) is given; a rung of the ladder is asked.

ax^2+bx+a=0 \;\Rightarrow\; x+\frac1x = \frac ba

Divide the whole equation by $x$ (legal: $x=0$ is not a root since $a\ne0$ ).
Read off $x+\frac1x = \frac ba$ .
Climb the ladder as usual.

Why: dividing by $x$ merges the two end terms into $a(x+\frac1x)$ .

Example: If $4x^2-9x+4=0$ , then $x^2+\frac1{x^2}$ equals:

$x+\frac1x = \frac94$ → $x^2+\frac1{x^2} = \frac{81}{16}-2 = \frac{49}{16}$ . No quadratic formula needed.

Type 4: Special / cyclic valuescommon2 practice Q

How to spot it:

$k$ is $1$ , $-1$ , $\sqrt3$ (or the asked expression has huge exponents like $x^{18}$ ); a small value is expected.

k=1 \Rightarrow x^3=-1; \quad k=-1 \Rightarrow x^3=1; \quad \text{reduce exponents mod } 3

From $x+\frac1x = \pm1$ , multiply through by $x$ : $x^2\pm x+1 = 0$ .
Multiply by $(x\mp1)$ : $(x\mp1)(x^2\pm x+1) = x^3\mp1 = 0$ → $x^3 = \pm1$ .
Reduce every exponent modulo 3 (or 6) and substitute.
For $k=\sqrt3$ : $x^2+\frac1{x^2} = -1$ , and higher even rungs alternate small values.

Why: these $x$ are cube roots of unity family — powers cycle with period 3.

Example: If $x+\frac1x=-1$ , then $x^6+x^3+1$ equals:

$x^2+x+1=0$ → $x^3 = 1$ → $x^6+x^3+1 = 1+1+1 = 3$ .

Formulas

Square

x^2+\frac{1}{x^2}=\left(x+\frac1x\right)^2-2=\left(x-\frac1x\right)^2+2

Cube (sum)

x^3+\frac{1}{x^3}=\left(x+\frac1x\right)^3-3\left(x+\frac1x\right)

if $x+\frac1x=k$ then it is $k^3-3k$

Cube (difference)

x^3-\frac{1}{x^3}=\left(x-\frac1x\right)^3+3\left(x-\frac1x\right)

Fourth power

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

Fifth power

x^5+\frac{1}{x^5}=\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)-\left(x+\frac1x\right)

Link between sum and difference

\left(x+\frac1x\right)^2-\left(x-\frac1x\right)^2=4

Quadratic to reciprocal form

ax^2-bx+a=0\ \Rightarrow\ x+\frac1x=\frac ba

Mixed form

\left(px+\frac{1}{qx}\right)^2=p^2x^2+\frac{1}{q^2x^2}+\frac{2p}{q}

Shortcut tricks

⚡ The ladder

From $k=x+\frac1x$ : $x^2+\frac{1}{x^2}=k^2-2$ , $x^4+\frac{1}{x^4}=(k^2-2)^2-2$ , $x^8+\frac{1}{x^8}=\ldots$ — keep squaring and subtracting 2. For the cube use $k^3-3k$ .

Example: If $x+\frac1x=4$ , find $x^2+\frac{1}{x^2}$ , $x^4+\frac{1}{x^4}$ and $x^3+\frac{1}{x^3}$ .

$16-2=14$ ; $14^2-2=194$ ; $4^3-3(4)=64-12=52$ .

⚡ Special values give cyclic powers

If $x+\frac1x=$	Then
$2$	$x=1$
$-2$	$x=-1$
$1$	$x^3=-1$ (so $x^6=1$ )
$-1$	$x^3=1$
$\sqrt3$	$x^6=-1$ (so $x^{12}=1$ )
$\sqrt2$	$x^4=-1$ (so $x^8=1$ )

Reduce every exponent using the cycle.

Example: If $x+\frac1x=1$ , find $x^{30}+x^{24}+x^{18}+x^{12}+x^6+1$ .

$x+\frac1x=1\Rightarrow x^2-x+1=0\Rightarrow x^3=-1\Rightarrow x^6=1$ . Every term is a power of $x^6$ , so each equals 1. Sum $=6$ .

⚡ Divide the quadratic by $x$

$ax^2-bx+a=0$ (equal first and last coefficients) means $x+\frac1x=\frac ba$ . Spot equal end coefficients instantly.

Example: If $3x^2-7x+3=0$ , find $x^2+\frac{1}{x^2}$ .

Divide by $3x$ : $x+\frac1x=\frac73$ . Then $x^2+\frac1{x^2}=\frac{49}{9}-2=\frac{31}{9}$ .

⚡ Mixed form $px+\frac{1}{qx}$

Squaring gives a middle term $2\cdot px\cdot\frac{1}{qx}=\frac{2p}{q}$ , not 2.

Example: If $2x+\frac{1}{3x}=5$ , find $4x^2+\frac{1}{9x^2}$ .

Square: $4x^2+\frac{1}{9x^2}+2\cdot 2x\cdot\frac{1}{3x}=25\Rightarrow 4x^2+\frac{1}{9x^2}=25-\frac43=\frac{71}{3}$ .

Where students lose marks

Subtracting 2 instead of adding 2 when going from $x-\frac1x$ to $x^2+\frac{1}{x^2}$ .
Writing $x^3+\frac{1}{x^3}=k^3-3$ instead of $k^3-3k$ .
Taking only the positive root of $\left(x-\frac1x\right)^2$ when no condition like $x>1$ is given — check the options.
For $2x+\frac{1}{3x}$ type, using a middle term of 2 instead of $\frac{2p}{q}$ .

Practice sets — 14 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 10 questions

Suggested time 6 min · wrong answers go to your mistake notebook automatically.