SSC CGL Maths · Algebraic Identities and Simplification

Cubes and Higher Powers of x ± 1/x

Cubes of x ± 1/x, sixth and eighth powers built from squares and cubes, and the product rule that gives fifth and seventh powers.

All Algebraic Identities and Simplification notes

Why this matters

30 questions in the bank, and all six HARD questions of the chapter sit on this page. They are long only in arithmetic: each higher power comes from two smaller ones by one product rule. Write the small values (x + 1/x, x² + 1/x², x³ + 1/x³) in a row first, and every higher power is a product of two of them.

Concept 1 of 3: Cubing x + 1/x and x − 1/x

Cubing

x + \frac{1}{x}

gives

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

plus 3 times

x + \frac{1}{x}

, because

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

again. Take the extra away and the cube is done.

Definition

$x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right)$ . This is $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$ with $ab = 1$ .
$x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right)$ .
Another route, handy when $x^2 + \frac{1}{x^2}$ is known: $x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2} - 1\right)$ and $x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2} + 1\right)$ .
Terms that multiply to k: for $x + \frac{k}{x}$ , $x^3 + \frac{k^3}{x^3} = \left(x + \frac{k}{x}\right)^3 - 3k\left(x + \frac{k}{x}\right)$ .
Backwards: given $x^3 + \frac{1}{x^3}$ , put each option t into $t^3 - 3t$ and keep the one that matches. For positive x, t is at least 2.
If you are given $x^2 + \frac{1}{x^2}$ or $x^2 - \frac{1}{x^2}$ , first climb down to $x \pm \frac{1}{x}$ with the ladder and the sign rule from the previous page.

Cubes of x ± 1/x

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

Worked example

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

, find

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 18 Sep 2025, Shift 3 · Q57Easy

Example 1 · Algebraic Identities and Simplification · Cubes and Higher Powers of x ± 1/x

\left(x + \frac{2}{x} = 7\right)

, find the value of

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

Minus 3 times what

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

, never

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

. With

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

the answer is 52, not 61.

The sign in the difference cube

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

: plus 3 times, not minus. The sum form and the difference form have opposite corrections.

Terms that do not multiply to 1

For

x + \frac{5}{x}

the two terms multiply to 5, so the correction is

3 \times 5 \times \left(x + \frac{5}{x}\right)

. Using 3 instead of 15 gives a printed wrong option.

Drill 9 more on cubing x + 1/x and x − 1/x

Concept 2 of 3: Sixth and eighth powers of x ± 1/x

An even power is a square of a smaller one or a cube of a smaller one, so

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

can be reached two ways. A difference such as

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

keeps splitting as a difference of squares until only known pieces are left.

Definition

$x^6 + \frac{1}{x^6} = \left(x^3 + \frac{1}{x^3}\right)^2 - 2$ , or $\left(x^2 + \frac{1}{x^2}\right)^3 - 3\left(x^2 + \frac{1}{x^2}\right)$ . The second is quicker when you start from $x - \frac{1}{x}$ , since $x^2 + \frac{1}{x^2}$ comes first.
$x^6 - \frac{1}{x^6} = \left(x^3 + \frac{1}{x^3}\right)\left(x^3 - \frac{1}{x^3}\right)$ , or $\left(x^2 - \frac{1}{x^2}\right)\left(x^4 + 1 + \frac{1}{x^4}\right)$ .
$x^8 - \frac{1}{x^8} = \left(x^4 + \frac{1}{x^4}\right)\left(x^4 - \frac{1}{x^4}\right) = \left(x^4 + \frac{1}{x^4}\right)\left(x^2 + \frac{1}{x^2}\right)\left(x^2 - \frac{1}{x^2}\right)$ .
A mixed expression such as $x^6 + \frac{1}{x^6} - 2\left(x^2 + \frac{1}{x^2}\right)$ is each rung worked out, then added. Write the rungs in a row first.
The sign: a difference of powers has the sign of $x - \frac{1}{x}$ . With $x > 1$ it is positive.

Sixth and eighth powers

x^6 + \frac{1}{x^6} = \left(x^3 + \frac{1}{x^3}\right)^2 - 2 \qquad x^6 + \frac{1}{x^6} = \left(x^2 + \frac{1}{x^2}\right)^3 - 3\left(x^2 + \frac{1}{x^2}\right) \qquad x^8 - \frac{1}{x^8} = \left(x^4 + \frac{1}{x^4}\right)\left(x^2 + \frac{1}{x^2}\right)\left(x^2 - \frac{1}{x^2}\right)

Worked example

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

, find

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

by both routes.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 18 Sep 2025, Shift 2 · Q56Hard

Example 2 · Algebraic Identities and Simplification · Cubes and Higher Powers of x ± 1/x

Given,

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

, Find the value of:

x^6 + \frac{1}{x^6} - 3\left(x^4 + \frac{1}{x^4}\right) + 4\left(x^2 + \frac{1}{x^2}\right)

The wrong correction when cubing a square

x^6 + \frac{1}{x^6} = \left(x^2 + \frac{1}{x^2}\right)^3 - 3\left(x^2 + \frac{1}{x^2}\right)

. The correction uses

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

, the thing being cubed, not

x + \frac{1}{x}

A missing factor in an eighth-power difference

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

splits into three known pieces:

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

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

and

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

. Stopping after the first two leaves out the difference, which carries the surd.

Drill 14 more on sixth and eighth powers of x ± 1/x

Concept 3 of 3: Fifth and seventh powers by multiplying two smaller ones

Multiply

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

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

and you get

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

plus two cross terms,

x + \frac{1}{x}

. The powers add on the outside and subtract on the inside, so take the inside pair away.

Definition

$\left(x^m + \frac{1}{x^m}\right)\left(x^n + \frac{1}{x^n}\right) = \left(x^{m+n} + \frac{1}{x^{m+n}}\right) + \left(x^{m-n} + \frac{1}{x^{m-n}}\right)$ .
So $x^5 + \frac{1}{x^5} = \left(x^2 + \frac{1}{x^2}\right)\left(x^3 + \frac{1}{x^3}\right) - \left(x + \frac{1}{x}\right)$ , and $x^7 + \frac{1}{x^7} = \left(x^3 + \frac{1}{x^3}\right)\left(x^4 + \frac{1}{x^4}\right) - \left(x + \frac{1}{x}\right)$ .
For a difference: $x^5 - \frac{1}{x^5} = \left(x^2 + \frac{1}{x^2}\right)\left(x^3 - \frac{1}{x^3}\right) - \left(x - \frac{1}{x}\right)$ .
Check two ways: $x^5 + \frac{1}{x^5}$ is also $\left(x + \frac{1}{x}\right)\left(x^4 + \frac{1}{x^4}\right) - \left(x^3 + \frac{1}{x^3}\right)$ . If both agree, the arithmetic is right.
The options often differ only in the last digits, so do the final subtraction carefully: the unit digit alone usually settles it.

Product rule for powers of x ± 1/x

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

Worked example

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

, find

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2023 · Tier I, 17 Jul 2023, Shift 2 · Q54Moderate

Example 3 · Algebraic Identities and Simplification · Cubes and Higher Powers of x ± 1/x

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

, then

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

is equal to:

Taking away the wrong rung

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

, the extra pair is

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

, because

3 - 2 = 1

. Take away the power equal to the difference of the two you multiplied, not one of them.

Drill 4 more on fifth and seventh powers by multiplying two smaller ones

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Cubing x + 1/x and x − 1/x
Cubes of x ± 1/x
$x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) \qquad x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right) \qquad x^3 + \frac{k^3}{x^3} = \left(x + \frac{k}{x}\right)^3 - 3k\left(x + \frac{k}{x}\right)$
Sixth and eighth powers of x ± 1/x
Sixth and eighth powers
$x^6 + \frac{1}{x^6} = \left(x^3 + \frac{1}{x^3}\right)^2 - 2 \qquad x^6 + \frac{1}{x^6} = \left(x^2 + \frac{1}{x^2}\right)^3 - 3\left(x^2 + \frac{1}{x^2}\right) \qquad x^8 - \frac{1}{x^8} = \left(x^4 + \frac{1}{x^4}\right)\left(x^2 + \frac{1}{x^2}\right)\left(x^2 - \frac{1}{x^2}\right)$
Fifth and seventh powers by multiplying two smaller ones
Product rule for powers of x ± 1/x
$\left(x^m + \frac{1}{x^m}\right)\left(x^n + \frac{1}{x^n}\right) = x^{m+n} + \frac{1}{x^{m+n}} + x^{m-n} + \frac{1}{x^{m-n}} \qquad x^5 + \frac{1}{x^5} = \left(x^2 + \frac{1}{x^2}\right)\left(x^3 + \frac{1}{x^3}\right) - \left(x + \frac{1}{x}\right)$

Watch out for (6)

Minus 3 times what
→ Cubing x + 1/x and x − 1/x
The sign in the difference cube
→ Cubing x + 1/x and x − 1/x
Terms that do not multiply to 1
→ Cubing x + 1/x and x − 1/x
The wrong correction when cubing a square
→ Sixth and eighth powers of x ± 1/x
A missing factor in an eighth-power difference
→ Sixth and eighth powers of x ± 1/x
Taking away the wrong rung
→ Fifth and seventh powers by multiplying two smaller ones

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