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

$\left( x+\dfrac{1}{3} \right)\left( x^2-\dfrac{1}{3}x+\dfrac{1}{9} \right)$
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, $x^3+\dfrac{1}{27} ,$ is \begin{array}{l} \left( x+\dfrac{1}{3} \right)\left[ (x)^2-(x)\left( \dfrac{1}{3} \right)+\left( \dfrac{1}{3} \right)^2 \right] \\\\= \left( x+\dfrac{1}{3} \right)\left( x^2-\dfrac{1}{3}x+\dfrac{1}{9} \right) .\end{array}