## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

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