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

Mar 10, 2016

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

#### 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:

${z}^{6} - 64 {p}^{6}$

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

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

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

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