# How do you factor 1-x^8 ?

Nov 15, 2016

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

#### Explanation:

Remember $\left({a}^{2} - {b}^{2}\right) = \left(a - b\right) \left(a + b\right)$
$\left(1 - {x}^{8}\right) = \left(1 - {x}^{4}\right) \left(1 + {x}^{4}\right)$
=$\left(1 - {x}^{2}\right) \left(1 + {x}^{2}\right) \left(1 + {x}^{4}\right)$
=$\left(1 - x\right) \left(1 + x\right) \left(1 + {x}^{2}\right) \left(1 + {x}^{4}\right)$

Nov 15, 2016

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

#### Explanation:

The difference of squares identity can be written:

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

Note also that:

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

In particular, putting $k = \sqrt{2}$ we find:

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

So we find:

$1 - {x}^{8} = {1}^{2} - {\left({x}^{4}\right)}^{2}$

$\textcolor{w h i t e}{1 - {x}^{8}} = \left(1 - {x}^{4}\right) \left(1 + {x}^{4}\right)$

$\textcolor{w h i t e}{1 - {x}^{8}} = \left({1}^{2} - {\left({x}^{2}\right)}^{2}\right) \left({1}^{4} + {x}^{4}\right)$

$\textcolor{w h i t e}{1 - {x}^{8}} = \left(1 - {x}^{2}\right) \left(1 + {x}^{2}\right) \left({1}^{2} - \sqrt{2} x + {x}^{2}\right) \left({1}^{2} + \sqrt{2} x + {x}^{2}\right)$

$\textcolor{w h i t e}{1 - {x}^{8}} = \left({1}^{2} - {x}^{2}\right) \left(1 + {x}^{2}\right) \left(1 - \sqrt{2} x + {x}^{2}\right) \left(1 + \sqrt{2} x + {x}^{2}\right)$

$\textcolor{w h i t e}{1 - {x}^{8}} = \left(1 - x\right) \left(1 + x\right) \left(1 + {x}^{2}\right) \left(1 - \sqrt{2} x + {x}^{2}\right) \left(1 + \sqrt{2} x + {x}^{2}\right)$

The remaining quadratic factors have no linear factors with Real coefficients.