SSC CGL Maths · Algebraic Identities and Simplification

The Cube Identity and a + b + c = 0

a³ + b³ + c³ − 3abc has a + b + c as a factor, so when a + b + c = 0 the sum of the three cubes is 3abc.

All Algebraic Identities and Simplification notes

Why this matters

26 questions in the bank, and seven of them come from the 2025 shifts. About half give two of the sums and ask for the cube expression; the other half hide three numbers that add to zero. Before cubing any number, add the bases: if they sum to zero, the cubes need no arithmetic at all.

Concept 1 of 2: The factorisation of a³ + b³ + c³ − 3abc

The expression is the sum

a + b + c

times the bracket

a^2 + b^2 + c^2 - ab - bc - ca

from the three-variable page. So it is built from sums you already know how to find.

Definition

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$ .
The bracket is half the sum of squared differences, so $a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a + b + c)\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right]$ .
With the sum $s = a + b + c$ and the pair sum $q = ab + bc + ca$ : $a^3 + b^3 + c^3 - 3abc = s(s^2 - 3q)$ .
For $a^3 + b^3 + c^3$ itself, add $3abc$ back: $a^3 + b^3 + c^3 = s^3 - 3sq + 3abc$ .
A minus sign: $a^3 + b^3 - c^3 + 3abc$ is $a^3 + b^3 + (-c)^3 - 3ab(-c)$ , so its factor is $a + b - c$ and its bracket is $a^2 + b^2 + c^2 - ab + bc + ca$ .
Number fractions: when the top is $a^3 + b^3 + c^3 - 3abc$ and the bottom is $a^2 + b^2 + c^2 - ab - bc - ca$ for the same three numbers, the value is $a + b + c$ .

The three-cube identity

a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \qquad a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a + b + c)\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right] \qquad a^3 + b^3 + c^3 = s^3 - 3sq + 3abc

sthe sum a + b + c
qthe sum of pair products ab + bc + ca

Worked example

a + b + c = 10

and

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

, find

a^3 + b^3 + c^3 - 3abc

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2023 · Tier I, 19 Jul 2023, Shift 4 · Q71Moderate

Example 1 · Algebraic Identities and Simplification · The Cube Identity and a + b + c = 0

(a + b + c) = 20

, and

a^2 + b^2 + c^2 = 152

, find the value of

a^3 + b^3 + c^3 - 3abc

Forgetting to add 3abc back

The identity gives

a^3 + b^3 + c^3 - 3abc

. If the question asks for

a^3 + b^3 + c^3

, add

3abc

at the end.

The bracket uses the sum of squares

The second factor is

a^2 + b^2 + c^2 - (ab + bc + ca)

. Putting

(a + b + c)^2

in place of

a^2 + b^2 + c^2

adds

2(ab + bc + ca)

too much.

Drill 12 more on the factorisation of a³ + b³ + c³ − 3abc

Concept 2 of 2: When a + b + c = 0, a³ + b³ + c³ = 3abc

a + b + c = 0

, the first factor of the identity is zero, so

a^3 + b^3 + c^3 - 3abc

is zero. The sum of the three cubes is then just three times their product.

Definition

$a + b + c = 0 \Rightarrow a^3 + b^3 + c^3 = 3abc$ .
Hidden zero sums: $x - y$ , $y - z$ and $z - x$ always add to zero, so $(x - y)^3 + (y - z)^3 + (z - x)^3 = 3(x - y)(y - z)(z - x)$ for any x, y, z.
A minus sign: $a + b = c$ means $a + b + (-c) = 0$ , so $a^3 + b^3 - c^3 = -3abc$ .
Coefficients: $27a^3 + b^3 + 8c^3 = 18abc$ is $(3a)^3 + b^3 + (2c)^3 = 3(3a)(b)(2c)$ , which holds when $3a + b + 2c = 0$ .
Two terms: if $a + b = 0$ then $b = -a$ and $a^3 + b^3 = 0$ .
In numbers: for $15^3 + 8^3 - 23^3$ , the bases $15 + 8 - 23 = 0$ , so the value is $3 \times 15 \times 8 \times (-23) = -8280$ .
Add the bases first. If they add to 7, not 0, the shortcut does not apply: cube and add, or use the full identity.

Three cubes when the sum is zero

a + b + c = 0 \Rightarrow a^3 + b^3 + c^3 = 3abc

Worked example

Find

\frac{(0.4)^3 + (0.5)^3 - (0.9)^3}{0.4 \times 0.5 \times 0.9}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 17 Sep 2025, Shift 3 · Q71Moderate

Example 2 · Algebraic Identities and Simplification · The Cube Identity and a + b + c = 0

19^3 + 20^3 - 39^3 + 118

is equal to:

Equal values also satisfy the cube condition

x^3 + y^3 + z^3 = 3xyz

also holds when

x = y = z

, because then the bracket is zero. So the condition gives

x + y + z = 0

x = y = z

: check the options for both.

A sum that is nearly zero

21^3 + 9^3 - 31^3

, the bases add to

-1

, not 0, so

3abc

is not the answer. Add the bases before using the shortcut.

Drill 12 more on when a + b + c = 0, a³ + b³ + c³ = 3abc

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 (2)

The factorisation of a³ + b³ + c³ − 3abc
The three-cube identity
$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \qquad a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a + b + c)\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right] \qquad a^3 + b^3 + c^3 = s^3 - 3sq + 3abc$
When a + b + c = 0, a³ + b³ + c³ = 3abc
Three cubes when the sum is zero
$a + b + c = 0 \Rightarrow a^3 + b^3 + c^3 = 3abc$

Watch out for (4)

Forgetting to add 3abc back
→ The factorisation of a³ + b³ + c³ − 3abc
The bracket uses the sum of squares
→ The factorisation of a³ + b³ + c³ − 3abc
Equal values also satisfy the cube condition
→ When a + b + c = 0, a³ + b³ + c³ = 3abc
A sum that is nearly zero
→ When a + b + c = 0, a³ + b³ + c³ = 3abc

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