# How do you factor x^6 - 1?

Jul 2, 2015

${x}^{6} - 1 = \left({x}^{3} - 1\right) \left({x}^{3} + 1\right)$

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

#### Explanation:

Use difference of squares, difference of cubes and sum of cubes:

[1]: ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

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

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

${x}^{6} - 1$

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

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

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

$= \left(x - 1\right) \left({x}^{2} + x + 1\right) \left(x + 1\right) \left({x}^{2} - x + 1\right)$ by [2] and [3]