# How do you factor 27w^3z-z^4w^6?

Apr 7, 2017

$27 {w}^{3} z - {z}^{4} {w}^{6} = {w}^{3} z \left(3 - z w\right) \left(9 + 3 z w + {z}^{2} {w}^{2}\right)$

#### 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 will use this with $a = 3$ and $b = z w$.

$\textcolor{w h i t e}{}$
Given:

$27 {w}^{3} z - {z}^{4} {w}^{6}$

First note that both of the terms are divisible by ${w}^{3}$ and by $z$, so by ${w}^{3} z$. So we can separate that out as a factor first:

$27 {w}^{3} z - {z}^{4} {w}^{6} = {w}^{3} z \left(27 - {z}^{3} {w}^{3}\right)$

$\textcolor{w h i t e}{27 {w}^{3} z - {z}^{4} {w}^{6}} = {w}^{3} z \left({3}^{3} - {\left(z w\right)}^{3}\right)$

$\textcolor{w h i t e}{27 {w}^{3} z - {z}^{4} {w}^{6}} = {w}^{3} z \left(3 - z w\right) \left({3}^{2} + 3 z w + {\left(z w\right)}^{2}\right)$

$\textcolor{w h i t e}{27 {w}^{3} z - {z}^{4} {w}^{6}} = {w}^{3} z \left(3 - z w\right) \left(9 + 3 z w + {z}^{2} {w}^{2}\right)$

This is as far as we can go with Real coefficients.

The remaining quartic factor $\left(9 + 3 z w + {z}^{2} {w}^{2}\right)$ can be factored further, but only with the use of Complex coefficients.