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

Dec 5, 2015

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

#### Explanation:

The sum of cubes formula states that

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

(Verify this by multiplying it out)

As $\frac{1}{8} = {\left(\frac{1}{2}\right)}^{3}$ we can apply the formula.

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

Note that we cannot factor any further, as the discriminant for the remaining quadratic is negative:
${\left(- \frac{1}{2}\right)}^{2} - 4 \left(1\right) \left(\frac{1}{4}\right) = \frac{1}{4} - 1 = - \frac{3}{4}$
and thus it has no real roots.