# How do you factor completely 7x^3-56 ?

Sep 25, 2016

$7 {x}^{3} - 56 = 7 \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$

#### Explanation:

The difference of cubes identity can be written:

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

We use this with $a = x$ and $b = 2$ below.

First note that both terms are divisible by $7$, so we can separate that out as a factor first:

$7 {x}^{3} - 56 = 7 \left({x}^{3} - 8\right)$

$\textcolor{w h i t e}{7 {x}^{3} - 56} = 7 \left({x}^{3} - {2}^{3}\right)$

$\textcolor{w h i t e}{7 {x}^{3} - 56} = 7 \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$

This is as far as we can go with Real coefficients. If we allow Complex coefficients, then this factors further as:

$\textcolor{w h i t e}{7 {x}^{3} - 56} = 7 \left(x - 2\right) \left(x - 2 \omega\right) \left(x - 2 {\omega}^{2}\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.