Chapter 5 - Polynomials and Factoring - 5.5 Factoring Sums or Differences of Cubes - 5.5 Exercise Set - Page 338: 23

$\left( a+\dfrac{1}{2} \right)\left( a^2-\dfrac{1}{2}a+\dfrac{1}{4} \right)$

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, $ a^3+\dfrac{1}{8} ,$ is \begin{array}{l} \left( a+\dfrac{1}{2} \right)\left[ (a)^2-(a)\left( \dfrac{1}{2} \right)+\left( \dfrac{1}{2} \right)^2 \right] \\\\= \left( a+\dfrac{1}{2} \right)\left( a^2-\dfrac{1}{2}a+\dfrac{1}{4} \right) .\end{array}