Chapter 5 - Polynomials and Factoring - 5.5 Factoring Sums or Differences of Cubes - 5.5 Exercise Set - Page 338: 24

$\left( x+\dfrac{1}{3} \right)\left( x^2-\dfrac{1}{3}x+\dfrac{1}{9} \right)$

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}