# How do you factor 125x^6 - y^6?

##### 1 Answer
Aug 26, 2016

$125 {x}^{6} - {y}^{6} = \left(\sqrt{5} x - y\right) \left(5 {x}^{2} + \sqrt{5} x y + {y}^{2}\right) \left(\sqrt{5} x + y\right) \left(5 {x}^{2} - \sqrt{5} x y + {y}^{2}\right)$

#### Explanation:

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

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

The sum of cubes identity can be written:

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

Hence we find:

$125 {x}^{6} - {y}^{6}$

$= {\left(\sqrt{125} {x}^{3}\right)}^{2} - {\left({y}^{3}\right)}^{2}$

$= \left(\sqrt{125} {x}^{3} - {y}^{3}\right) \left(\sqrt{125} {x}^{3} + {y}^{3}\right)$

$= \left({\left(\sqrt{5} x\right)}^{3} - {y}^{3}\right) \left({\left(\sqrt{5} x\right)}^{3} + {y}^{3}\right)$

$= \left(\sqrt{5} x - y\right) \left({\left(\sqrt{5} x\right)}^{2} + \left(\sqrt{5} x\right) y + {y}^{2}\right) \left(\sqrt{5} x + y\right) \left({\left(\sqrt{5} x\right)}^{2} - \left(\sqrt{5} x\right) y + {y}^{2}\right)$

$= \left(\sqrt{5} x - y\right) \left(5 {x}^{2} + \sqrt{5} x y + {y}^{2}\right) \left(\sqrt{5} x + y\right) \left(5 {x}^{2} - \sqrt{5} x y + {y}^{2}\right)$

There are no simpler factors with Real coefficients.

If you allow Complex coefficients then you can factor this further as:

$= \left(\sqrt{5} x - y\right) \left(\sqrt{5} x - \omega y\right) \left(\sqrt{5} x - {\omega}^{2} y\right) \left(\sqrt{5} x + y\right) \left(\sqrt{5} x + \omega y\right) \left(\sqrt{5} x + {\omega}^{2} y\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.