How do you factor 64x^6 -1?

Jan 19, 2016

Use some standard identities to find:

$64 {x}^{6} - 1 = \left(2 x - 1\right) \left(4 {x}^{2} + 2 x + 1\right) \left(2 x + 1\right) \left(4 {x}^{2} - 2 x + 1\right)$

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:

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

We find:

$64 {x}^{6} - 1$

$= {\left(8 {x}^{3}\right)}^{2} - {1}^{2}$

$= \left(8 {x}^{3} - 1\right) \left(8 {x}^{3} + 1\right)$

$= \left({\left(2 x\right)}^{3} - {1}^{3}\right) \left({\left(2 x\right)}^{3} + {1}^{3}\right)$

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

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

This is as far as we can go with Real coefficients, but if you allow Complex coefficients then this factors further as:

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

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