# How do you factor 4x^3-108?

May 25, 2016

$4 \left(x - 3\right) \left({x}^{2} + 3 x + 9\right)$

#### Explanation:

To factor $4 {x}^{3} - 108$

Begin by factoring out a $4$

$4 \left({x}^{3} - 27\right)$

$\left({x}^{3} - 27\right)$ is a binomial of perfect cubes.

This factors as

${a}^{3} - {b}^{3}$

$\left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

Same Sign first
Opposite Sign next
then Always Positive

The cube of ${\sqrt[3]{x}}^{3}$ = x
The cube of $\sqrt[3]{27} = 3$

$\left(x - 3\right) \left({x}^{2} + 3 x + 9\right)$

The complete factorization is

$4 \left(x - 3\right) \left({x}^{2} + 3 x + 9\right)$