# How do you factor x^3 - 16?

Apr 7, 2016

${x}^{3} - 16$

$= \left(x - 2 \sqrt[3]{2}\right) \left({x}^{2} + 2 \sqrt[3]{2} x + 4 \sqrt[3]{4}\right)$

$= \left(x - 2 \sqrt[3]{2}\right) \left(x - 2 \sqrt[3]{2} \omega\right) \left(x - 2 \sqrt[3]{2} {\omega}^{2}\right)$

#### Explanation:

Use the difference of cubes identity:

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

with $a = x$ and $b = 2 \sqrt[3]{2}$

So:

${x}^{3} - 16 = {x}^{3} - {\left(2 \sqrt[3]{2}\right)}^{3}$

$= \left(x - 2 \sqrt[3]{2}\right) \left({x}^{2} + 2 \sqrt[3]{2} x + 4 \sqrt[3]{4}\right)$

This is as far as we can go with Real coefficients, but if you allow Complex coefficients then this can be further factored as:

$= \left(x - 2 \sqrt[3]{2}\right) \left(x - 2 \sqrt[3]{2} \omega\right) \left(x - 2 \sqrt[3]{2} {\omega}^{2}\right)$

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