SSC CGL Maths · Algebraic Identities and Simplification

Cubes of a Binomial

The cube of a sum or a difference, the sum and difference of two cubes, and the number fractions those factors cancel.

All Algebraic Identities and Simplification notes

Why this matters

31 questions in the bank, and almost half of them are MODERATE. Three shapes repeat: two numbers given by a sum or difference and a product, an expanded cube to spot, and a fraction whose top is a sum or difference of cubes and whose bottom is the matching quadratic. The fractions look like heavy arithmetic and take five seconds once you see the pattern.

Concept 1 of 3: Cubes from a sum or difference and a product

(a + b)^3

holds

a^3 + b^3

plus extra terms that are all multiples of

ab(a + b)

. Take those away and the sum of cubes is left, with no need to find a and b.

Definition

$(a + b)^3 = a^3 + b^3 + 3ab(a + b)$ , so $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$ .
$(a - b)^3 = a^3 - b^3 - 3ab(a - b)$ , so $a^3 - b^3 = (a - b)^3 + 3ab(a - b)$ .
Backwards: given $a^3 + b^3$ and $a + b$ , the same line gives $ab$ . Given $a^3 + b^3$ and $ab$ , try each option for $a + b$ in $s^3 - 3ab\,s$ .
If the question needs the other sign ( $a - b$ from $a + b$ ), go through the square first: $(a - b)^2 = (a + b)^2 - 4ab$ .
A coefficient rides along: for $3x + y$ and $xy$ , take $a = 3x$ , $b = y$ and $ab = 3xy$ .
Fast route: with natural numbers, find the pair. A difference of 2 and a product of 15 are 5 and 3; then cube them. Use the identity when no whole pair fits.

Cube identities for two terms

(a + b)^3 = a^3 + b^3 + 3ab(a + b) \qquad (a - b)^3 = a^3 - b^3 - 3ab(a - b) \qquad a^3 + b^3 = (a + b)^3 - 3ab(a + b) \qquad a^3 - b^3 = (a - b)^3 + 3ab(a - b)

Worked example

x + y = 9

and

xy = 14

, find

x^3 + y^3

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 19 Sep 2025, Shift 2 · Q57Moderate

Example 1 · Algebraic Identities and Simplification · Cubes of a Binomial

a^3 + b^3=35

and

ab = 6

, find the value of

a + b

The correction carries the sum

a^3 + b^3 = (a + b)^3 - 3ab(a + b)

, not

(a + b)^3 - 3ab

. Leaving out the factor

(a + b)

is the commonest wrong option.

The sign in the difference form

a^3 - b^3 = (a - b)^3 + 3ab(a - b)

: plus, because

(a - b)^3

already took

3ab(a - b)

away. Writing minus gives a number that is too small.

Drill 13 more on cubes from a sum or difference and a product

Concept 2 of 3: Spotting an expanded cube

Four terms whose first and last are cubes, with middle coefficients that follow the 1, 3, 3, 1 pattern, are a cube in disguise. Spotting it turns a long substitution into one small number cubed.

Definition

$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ and $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ .
The test: the first and last terms are $a^3$ and $b^3$ . Then the middle terms must be exactly $3a^2b$ and $3ab^2$ . For $8x^3 + 36x^2y + 54xy^2 + 27y^3$ : $a = 2x$ , $b = 3y$ , $3a^2b = 36x^2y$ and $3ab^2 = 54xy^2$ , so it is $(2x + 3y)^3$ .
Complete the cube: $a^3 + 3a^2 + 3a + 1 = (a + 1)^3$ , so add or take away the missing constant on both sides.
A near-cube is not a cube. If a middle coefficient does not match, write the expression as a cube plus a leftover, or put the numbers in term by term. SSC prints near-cubes on purpose.

Expanded cubes

(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \qquad (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3

Worked example

Find

8x^3 - 36x^2 + 54x - 27

when

x = 4

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 17 Sep 2024, Shift 2 · Q69Easy

Example 2 · Algebraic Identities and Simplification · Cubes of a Binomial

a^3 + 3a^2 + 3a = 7

, then the value of

a^2 + 2a

is:

Cubing a near-cube

27x^3 + 54x^2y + 36xy^2 + 8y^3

(3x + 2y)^3

, but change one middle coefficient and it is not. Check both middle coefficients before you cube; otherwise substitute term by term.

Drill 5 more on spotting an expanded cube

Concept 3 of 3: Sum and difference of cubes as factors

a^3 + b^3

always splits into

a + b

times a quadratic. When a fraction puts that quadratic underneath, it cancels and only

a + b

is left.

Definition

$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ and $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ .
So $\frac{a^3 + b^3}{a^2 - ab + b^2} = a + b$ and $\frac{a^3 - b^3}{a^2 + ab + b^2} = a - b$ . Turned over, they give $\frac{1}{a + b}$ and $\frac{1}{a - b}$ .
Pair by the signs: a plus between the cubes goes with a minus $ab$ in the quadratic; a minus between the cubes goes with a plus $ab$ .
The middle product may be printed as one number: 0.24 is $0.6 \times 0.4$ , and 0.16 is $0.4^2$ .
The same pair on top and bottom. Check that the numbers in the quadratic are the ones that are cubed before you cancel.
A product such as $\left(\frac{2}{x} + 2x\right)\left(\frac{4}{x^2} - 4 + 4x^2\right)$ is a sum of cubes: $a = \frac{2}{x}$ , $b = 2x$ , $ab = 4$ , so it equals $\frac{8}{x^3} + 8x^3$ .

Sum and difference of cubes

a^3 + b^3 = (a + b)(a^2 - ab + b^2) \qquad a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Worked example

Simplify

\frac{(0.9)^3 + (0.6)^3}{(0.9)^2 - 0.54 + (0.6)^2}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 11 Sep 2024, Shift 2 · Q66Easy

Example 3 · Algebraic Identities and Simplification · Cubes of a Binomial

Simplify

\frac{(4.2)^3 - (1.2)^3}{(4.2)^2 + 5.04 + (1.2)^2}

Pairing the wrong quadratic

a^3 + b^3

cancels with

a^2 - ab + b^2

, not with

a^2 + ab + b^2

. With the wrong pair nothing cancels and the value is not

a + b

Different numbers on top and below

The shortcut needs the same pair in both places. If the top cubes 0.2 and 0.3 but the bottom is built from 0.5 and 0.3, nothing cancels: work the arithmetic instead.

Drill 10 more on sum and difference of cubes as factors

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)

Cubes from a sum or difference and a product
Cube identities for two terms
$(a + b)^3 = a^3 + b^3 + 3ab(a + b) \qquad (a - b)^3 = a^3 - b^3 - 3ab(a - b) \qquad a^3 + b^3 = (a + b)^3 - 3ab(a + b) \qquad a^3 - b^3 = (a - b)^3 + 3ab(a - b)$
Spotting an expanded cube
Expanded cubes
$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \qquad (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$
Sum and difference of cubes as factors
Sum and difference of cubes
$a^3 + b^3 = (a + b)(a^2 - ab + b^2) \qquad a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Watch out for (5)

The correction carries the sum
→ Cubes from a sum or difference and a product
The sign in the difference form
→ Cubes from a sum or difference and a product
Cubing a near-cube
→ Spotting an expanded cube
Pairing the wrong quadratic
→ Sum and difference of cubes as factors
Different numbers on top and below
→ Sum and difference of cubes as factors

Test yourself on Algebraic Identities and Simplification

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