# How do you factor x^3-27?

Mar 15, 2016

Use the difference of cubes identity to find:

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

#### Explanation:

Both ${x}^{3}$ and $27 = {3}^{3}$ are perfect cubes. So we can 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 $b = 3$ as follows:

${x}^{3} - 27$

$= {x}^{3} - {3}^{3}$

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

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

This is as far as you can go with Real coefficients. If you allow Complex coefficients then you can factor this a little further:

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

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