CDS Mathematics · Algebraic Identities and Simplification

The Cube Identity and a + b + c = 0

a³ + b³ + c³ − 3abc factors as (a + b + c)(a² + b² + c² − ab − bc − ca), so when a + b + c = 0 the sum of the cubes is exactly 3abc.

All Algebraic Identities and Simplification notes CDS Mathematics strategy

Why this matters

Eleven PYQs, and the same fraction — (x − y)³ + (y − z)³ + (z − x)³ against (x − y)(y − z)(z − x) — has been set three times with different constants. Check that the three pieces add to zero; the sum of their cubes is then three times their product.

Concept 1 of 2: When three numbers add to zero

a + b + c = 0

, the factor

(a + b + c)

in the cube identity is zero, so

a^3 + b^3 + c^3 - 3abc = 0

. The three differences

x - y

y - z

z - x

always add to zero, which is why they keep appearing.

Definition

If $a + b + c = 0$ :

$a^3 + b^3 + c^3 = 3abc$ ;
$a^2 + b^2 + c^2 = -2(ab + bc + ca)$ ;
$b + c = -a$ , $c + a = -b$ , $a + b = -c$ , so brackets like $(b + c - a)$ become $-2a$ .

Always true: $(x - y) + (y - z) + (z - x) = 0$ .

Zero sum

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

Worked example

Evaluate

17^3 - 12^3 - 5^3

without cubing.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (I) 2019 — Elementary Mathematics · Q98Moderate

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

What is the value of

\frac{(x-y)^3 + (y-z)^3 + (z-x)^3}{9(x-y)(y-z)(z-x)}

Check that the three really add to zero

(x - y)

(y - z)

(z - x)

add to

0

;

(x - y)

(y - z)

(x - z)

do not. Read the third bracket's order before using the shortcut.

Drill 6 more on when three numbers add to zero

Concept 2 of 2: The full factorisation

The second factor of the cube identity is half the sum of the squares of the differences, so it is zero only when all three numbers are equal. That makes the whole expression zero in exactly two situations.

Definition

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$ .
The second factor is $(a + b + c)^2 - 3(ab + bc + ca)$ , and also $\dfrac12\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right]$ .
So $a^3 + b^3 + c^3 = 3abc$ exactly when $a + b + c = 0$ or $a = b = c$ .
With $a^2 - bc = \alpha$ and so on, $a\alpha + b\beta + c\gamma = a^3 + b^3 + c^3 - 3abc$ and $\alpha + \beta + \gamma$ is the second factor.

Cube identity

a^3 + b^3 + c^3 - 3abc = (a + b + c)\left[(a + b + c)^2 - 3(ab + bc + ca)\right]

Worked example

a + b + c = 6

and

ab + bc + ca = 11

, find

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2025 · CDS (I) 2025 — Elementary Mathematics · Q40Easy

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

If x, y, z are real numbers such that

x + y + z = 10

and

xy + yz + zx = 18

, then what is the value of

x^3 + y^3 + z^3 - 3xyz

Zero does not mean a + b + c = 0

a^3 + b^3 + c^3 - 3abc = 0

also holds when

a = b = c

. A statement claiming it forces the numbers to be equal is not enough;

(1, -1, 0)

satisfies it with unequal numbers.

Drill 3 more on the full factorisation

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)

When three numbers add to zero
Zero sum
$a + b + c = 0 \;\Rightarrow\; a^3 + b^3 + c^3 = 3abc$
The full factorisation
Cube identity
$a^3 + b^3 + c^3 - 3abc = (a + b + c)\left[(a + b + c)^2 - 3(ab + bc + ca)\right]$

Watch out for (2)

Check that the three really add to zero
→ When three numbers add to zero
Zero does not mean a + b + c = 0
→ The full factorisation

Test yourself on Algebraic Identities and Simplification

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Algebraic Identities and Simplification chapter test Drill all 11 questions