# How do you factor 27x^3-512?

Dec 17, 2015

Use the difference of cubes identity to find:

$27 {x}^{3} - 512 = \left(3 x - 8\right) \left(9 {x}^{2} + 24 x + 64\right)$

#### Explanation:

The difference of cubes identity may be written:

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

Notice that $27 {x}^{3} = {\left(3 x\right)}^{3}$ and $512 = {8}^{3}$ are both perfect cubes. So let $a = 3 x$ and $b = 8$ to find:

$27 {x}^{3} - 512$

$= {\left(3 x\right)}^{3} - {8}^{3}$

$= \left(3 x - 8\right) \left({\left(3 x\right)}^{2} + \left(3 x\right) \left(8\right) + {8}^{2}\right)$

$= \left(3 x - 8\right) \left(9 {x}^{2} + 24 x + 64\right)$

If you allow Complex coefficients then this factors a little further:

$= \left(3 x - 8\right) \left(3 x - 8 \omega\right) \left(3 x - 8 {\omega}^{2}\right)$

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