# How do you factor x^6-1?

Jan 2, 2016

Use some standard identities to find:

${x}^{6} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right) \left(x + 1\right) \left({x}^{2} - x + 1\right)$

#### Explanation:

Use the difference of squares identity:

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

the difference of cubes identity:

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

and the sum of cubes identity:

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

as follows:

${x}^{6} - 1$

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

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

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

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

That's as far as we can go with Real coefficients.

If we allow Complex coefficients then this factors further as:

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

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