Answer
$\left( x-\dfrac{1}{2} \right)\left( x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)$
Work Step by Step
Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, the factored form of the given expression, $
x^3-\dfrac{1}{8}
,$ is
\begin{array}{l}
\left( x-\dfrac{1}{2} \right)\left[ (x)^2-(x)\left( -\dfrac{1}{2} \right)+\left( -\dfrac{1}{2} \right)^2\right]
\\\\=
\left( x-\dfrac{1}{2} \right)\left( x^2+\dfrac{1}{2}x+\dfrac{1}{4}\right)
.\end{array}