# How do you factor completely x^3+1/8?

Mar 3, 2016

${x}^{3} + \frac{1}{8} = \left(x + \frac{1}{2}\right) \left({x}^{2} - \frac{1}{2} x + \frac{1}{4}\right)$

#### Explanation:

Both ${x}^{3}$ and $\frac{1}{8} = {\left(\frac{1}{2}\right)}^{3}$ are perfect cubes. So we can use the "sum of cubes" identity:

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

with $a = x$ and $b = \frac{1}{2}$ as follows...

${x}^{3} + \frac{1}{8}$

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

$= \left(x + \frac{1}{2}\right) \left({x}^{2} - x \left(\frac{1}{2}\right) + {\left(\frac{1}{2}\right)}^{2}\right)$

$= \left(x + \frac{1}{2}\right) \left({x}^{2} - \frac{1}{2} x + \frac{1}{4}\right)$

The remaining quadratic factor has no simpler factors with Real coefficients, but we can factor it if we allow Complex coefficients:

$= \left(x + \frac{1}{2}\right) \left(x + \frac{1}{2} \omega\right) \left(x + \frac{1}{2} {\omega}^{2}\right)$

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