Identity

$a^{3} + b^{3} + c^{3} - 3 ab c$ , and the case $a + b + c = 0$

Algebra • Expressions • Algebraic Identities

Formula

a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)

For all

a

b

c

a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)

; in particular, if

a + b + c = 0

then

a^{3} + b^{3} + c^{3} = 3 ab c

How to use it

1Whenever three quantities that appear cubed add to zero, replace the sum of their cubes by three times their product.
2Differences such as $x - y$ , $y - z$ and $z - x$ add to zero identically, so the collapse applies to them without any condition on $x, y, z$ .
3Read the identity from right to left to factorise: $a^{3} + b^{3} + c^{3} - 3 ab c$ always carries $a + b + c$ as a factor.

Evaluate

7^{3} + (- 3)^{3} + (- 4)^{3}

without cubing each term.

Step-by-step

1
$7 + (- 3) + (- 4) = 0$
Why: Check the condition before using the collapsed form.
2
$a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3 ab c$
3
$3 \times 7 \times (- 3) \times (- 4) = 252$

Answer

252

Watch out

The collapse needs the three terms to sum to exactly zero; otherwise the factor

(a + b + c)

survives and

a^{3} + b^{3} + c^{3}

exceeds

3 ab c

(a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)

Builds on

Also known as

sum of three cubes, a^3+b^3+c^3-3abc, if a+b+c=0 then a^3+b^3+c^3=3abc, three cubes identity, symmetric cubic factorisation