# How do you factor the expression x^8 - 1?

Dec 20, 2015

Use some identities to find:

${x}^{8} - 1 = \left(x - 1\right) \left(x + 1\right) \left({x}^{2} + 1\right) \left({x}^{2} + \sqrt{2} x + 1\right) \left({x}^{2} - \sqrt{2} x + 1\right)$

#### Explanation:

We shall use some identities to help:

The difference of squares identity:

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

Consider also:

$\left({a}^{2} + k a b + {b}^{2}\right) \left({a}^{2} - k a b + {b}^{2}\right) = {a}^{4} + \left(2 - {k}^{2}\right) {a}^{2} {b}^{2} + {b}^{4}$

In particular if $k = \sqrt{2}$ then:

$\left({a}^{2} + \sqrt{2} a b + {b}^{2}\right) \left({a}^{2} - \sqrt{2} a b + {b}^{2}\right) = {a}^{4} + {b}^{4}$

So:

${x}^{8} - 1$

$= {\left({x}^{4}\right)}^{2} - {1}^{2}$

$= \left({x}^{4} - 1\right) \left({x}^{4} + 1\right)$

$= \left({\left({x}^{2}\right)}^{2} - {1}^{2}\right) \left({x}^{4} + 1\right)$

$= \left({x}^{2} - 1\right) \left({x}^{2} + 1\right) \left({x}^{4} + 1\right)$

$= \left({x}^{2} - {1}^{2}\right) \left({x}^{2} + 1\right) \left({x}^{4} + 1\right)$

$= \left(x - 1\right) \left(x + 1\right) \left({x}^{2} + 1\right) \left({x}^{4} + 1\right)$

$= \left(x - 1\right) \left(x + 1\right) \left({x}^{2} + 1\right) \left({x}^{4} + {1}^{4}\right)$

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

This is as far as we can go with Real coefficients. The remaining quadratic factors all have Complex zeros.