# How do you factor completely 64 - 27y^3 ?

Aug 5, 2016

$64 - 27 {y}^{3} = \left(4 - y\right) \left(16 + 4 y + {y}^{2}\right)$

#### Explanation:

The difference of cubes identity can be written:

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

Note that $64 = {4}^{3}$ and $27 {y}^{3} = {\left(3 y\right)}^{3}$ are both perfect cubes, so we find:

$64 - 27 {y}^{3}$

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

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

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

The remaining quadratic has negative discriminant. It cannot be factored further with Real coefficients. If we allow Complex coefficients then we can factor it completely:

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

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