# How do you factor 64r^3-p^3?

Apr 23, 2016

$64 {r}^{3} - {p}^{3} = \left(4 r - p\right) \left(16 {r}^{2} + 4 r p + {p}^{2}\right)$

#### Explanation:

The difference of cubes identity can be written:

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

We can use this with $a = 4 r$ and $b = p$ as follows:

$64 {r}^{3} - {p}^{3}$

$= {\left(4 r\right)}^{3} - {p}^{3}$

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

$= \left(4 r - p\right) \left(16 {r}^{2} + 4 r p + {p}^{2}\right)$

This cannot be factored further with Real coefficients.

If you allow Complex coeffients then it can be factored a little further:

$= \left(4 r - p\right) \left(4 r - \omega p\right) \left(4 r - {\omega}^{2} p\right)$

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