Identity

Cube of a sum and of a difference

Algebra • Expressions • Algebraic Identities

Formula

(a + b)^{3} = a^{3} + b^{3} + 3 ab (a + b)

For all

a

and

b

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

, and

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

How to use it

1When $a + b$ is itself the unknown, use the grouped form $a^{3} + b^{3} + 3 ab (a + b)$ : it turns the cube into an equation in $a + b$ and the individual terms never reappear.
2Cubing is worth doing when $a^{3}$ , $b^{3}$ and $ab$ are all rational while $a$ and $b$ are not - which is the usual situation with cube roots.
3Read it backwards to recognise $a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ as a perfect cube.

u = 5^{1/3} + 5^{2/3}

, express

u^{3}

in terms of

u

Step-by-step

1
$u^{3} = (5^{1/3})^{3} + (5^{2/3})^{3} + 3 \cdot 5^{1/3} \cdot 5^{2/3} (5^{1/3} + 5^{2/3})$
Why: Apply the grouped form with $a = 5^{1/3}$ and $b = 5^{2/3}$ .
2
$u^{3} = 5 + 25 + 3 \cdot 5 \cdot u = 30 + 15 u$
Why: $5^{1/3} \cdot 5^{2/3} = 5$ because the indices add to $1$ , and the bracket is $u$ itself.

Answer

u^{3} = 30 + 15 u

Watch out

The grouped form is the same identity, not a second one; it earns its place only when

a + b

is the quantity being solved for. And

(a + b)^{3}

is never

a^{3} + b^{3}

Builds on

Also known as

cube of a sum, binomial cube, (a+b)^3 expansion, cube of a binomial, a^3+b^3+3ab(a+b), cube both sides