# How do you factor completely m^8 +m^4/16 + 1/256?

Sep 17, 2016

${m}^{8} + {m}^{4} / 16 + \frac{1}{256}$

$= \left({m}^{2} - \frac{\sqrt{3}}{2} m + \frac{1}{4}\right) \left({m}^{2} + \frac{\sqrt{3}}{2} m + \frac{1}{4}\right) \left({m}^{2} - \frac{1}{2} m + \frac{1}{4}\right) \left({m}^{2} + \frac{1}{2} m + \frac{1}{4}\right)$

#### Explanation:

Note that:

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

Let $n = \frac{1}{2}$

Then:

${m}^{8} + {m}^{4} / 16 + \frac{1}{256}$

$= {\left({m}^{2}\right)}^{4} + {\left({m}^{2}\right)}^{2} {\left({n}^{2}\right)}^{2} + {\left({n}^{2}\right)}^{4}$

$= \left({\left({m}^{2}\right)}^{2} - \left({m}^{2}\right) \left({n}^{2}\right) + {\left({n}^{2}\right)}^{2}\right) \left({\left({m}^{2}\right)}^{2} + \left({m}^{2}\right) \left({n}^{2}\right) + {\left({n}^{2}\right)}^{2}\right)$

$= \left({m}^{4} - {m}^{2} {n}^{2} + {n}^{4}\right) \left({m}^{4} + {m}^{2} {n}^{2} + {n}^{4}\right)$

$= \left({m}^{2} - \sqrt{3} m n + {n}^{2}\right) \left({m}^{2} + \sqrt{3} m n + {n}^{2}\right) \left({m}^{2} - m n + {n}^{2}\right) \left({m}^{2} + m n + {n}^{2}\right)$

$= \left({m}^{2} - \frac{\sqrt{3}}{2} m + \frac{1}{4}\right) \left({m}^{2} + \frac{\sqrt{3}}{2} m + \frac{1}{4}\right) \left({m}^{2} - \frac{1}{2} m + \frac{1}{4}\right) \left({m}^{2} + \frac{1}{2} m + \frac{1}{4}\right)$