# How do you factor completely 27x^3-125y^3?

Apr 26, 2016

$27 {x}^{3} - 125 {y}^{3} = \left(3 x - 5 y\right) \left(9 {x}^{2} + 15 x y + 25 {y}^{2}\right)$

#### Explanation:

Notice that both of the terms are perfect cubes:

$27 {x}^{3} = {\left(3 x\right)}^{3}$

$125 {y}^{3} = {\left(5 y\right)}^{3}$

So we can conveniently use the difference of cubes identity:

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

with $A = 3 x$ and $B = 5 y$ as follows:

$27 {x}^{3} - 125 {y}^{3}$

$= {\left(3 x\right)}^{3} - {\left(5 y\right)}^{3}$

$= \left(3 x - 5 y\right) \left({\left(3 x\right)}^{2} + \left(3 x\right) \left(5 y\right) + {\left(5 y\right)}^{2}\right)$

$= \left(3 x - 5 y\right) \left(9 {x}^{2} + 15 x y + 25 {y}^{2}\right)$

The remaining quadratic factor has no linear factors with Real coefficients. If we allow Complex coefficients then we can factor a little further:

$= \left(3 x - 5 y\right) \left(3 x - 5 \omega y\right) \left(3 x - 5 {\omega}^{2} y\right)$

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