# How do you factor r^3 - 1?

Dec 20, 2015

Use the difference of cubes identity to find:

${r}^{3} - 1 = \left(r - 1\right) \left({r}^{2} + r + 1\right)$

#### Explanation:

The difference of cubes identity can be written:

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

Use this with $a = r$ and $b = 1$ as follows:

${r}^{3} - 1$

$= {r}^{3} - {1}^{3}$

$= \left(r - 1\right) \left({r}^{2} + \left(r\right) \left(1\right) + {1}^{2}\right)$

$= \left(r - 1\right) \left({r}^{2} + r + 1\right)$

If you allow Complex coefficients then this can be factored a little further:

$= \left(r - 1\right) \left(r - \omega\right) \left(r - {\omega}^{2}\right)$

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