# How do you factor x^8 - 1/81?

Dec 13, 2015

$\left({x}^{8} - \frac{1}{81}\right) = \left({x}^{2} - \frac{1}{3}\right) \left({x}^{2} + \frac{1}{3}\right) \left({x}^{4} + \frac{1}{9}\right)$

#### Explanation:

Given ${x}^{8} - \frac{1}{81}$

Using difference of square ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
$\implies {\left({x}^{4}\right)}^{2} - {\left(\frac{1}{9}\right)}^{2}$

$\implies \left({x}^{4} - \frac{1}{9}\right) \left({x}^{4} + \frac{1}{9}\right)$

Use the difference of square again...

$\implies \left({x}^{2} - \frac{1}{3}\right) \left({x}^{2} + \frac{1}{3}\right) \left({x}^{4} + \frac{1}{9}\right)$