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

Apr 9, 2015

This problem involves factoring the sum of cubes.

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

For the expression $\left({x}^{3} + \frac{1}{8}\right)$:

$a = \sqrt[3]{{x}^{3}} = x$

$b = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$

Solution:

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

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

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