# How do you factor  8x^3 - 216?

##### 1 Answer
Jun 24, 2016

$8 \left(x - 3\right) \left({x}^{2} + 3 x + 9\right)$

#### Explanation:

First step is to 'take out' a common factor of 8.

$\Rightarrow 8 \left({x}^{3} - 27\right) \ldots \ldots . . \left(A\right)$

now ${x}^{3} - 27 \text{ is a " color(blue)"difference of cubes}$ and is factorised as shown.

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

${x}^{3} = {\left(x\right)}^{3} \text{ and } 27 = {\left(3\right)}^{3}$

hence a = x and b = 3

$\Rightarrow {x}^{3} - 27 = \left(x - 3\right) \left({x}^{2} + 3 x + {3}^{2}\right) = \left(x - 3\right) \left({x}^{2} + 3 x + 9\right)$

Substituting this back into (A)

$\Rightarrow 8 {x}^{3} - 216 = 8 \left(x - 3\right) \left({x}^{2} + 3 x + 9\right)$