# How do you factor completely a^3 - 8?

Jan 4, 2016

${a}^{3} - 8 = \left(a - 2\right) \left({a}^{2} + 2 a + 4\right)$

$= \left(a - 2\right) \left(a + 1 - \sqrt{3} i\right) \left(a + 1 + \sqrt{3} i\right)$

#### Explanation:

Use the difference of cubes identity, which can be written:

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

Note that both ${a}^{3}$ and $8 = {2}^{3}$ are both perfect cubes.

So we find:

${a}^{3} - 8$

$= {a}^{3} - {2}^{3}$

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

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

The remaining quadratic factor has negative discriminant, so no Real zeros and no linear factors with Real coefficients. It is possible to factor it using Complex coefficients:

${a}^{2} + 2 a + 4$

$= {a}^{2} + 2 a + 1 + 3$

$= {\left(a + 1\right)}^{2} + 3$

$= {\left(a + 1\right)}^{2} - {\left(\sqrt{3} i\right)}^{2}$

$= \left(a + 1 - \sqrt{3} i\right) \left(a + 1 + \sqrt{3} i\right)$

So:

${a}^{8} - 8 = \left(a - 2\right) \left(a + 1 - \sqrt{3} i\right) \left(a + 1 + \sqrt{3} i\right)$

Another way to express the full factoring is:

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

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