# How do you factor completely 64-z^6?

Dec 20, 2015

Use difference of squares, difference of cubes and sum of cubes identities to find:

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

#### Explanation:

The difference of squares identity may be written:

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

The difference of cubes identity may be written:

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

The sum of cubes identity may be written:

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

So we find:

$64 - {z}^{6}$

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

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

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

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

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

The remaining quadratic factors have no simpler factors with Real coefficients, but if you allow Complex coefficients then you can go a little further:

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

or if you prefer:

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

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