Chapter 5 - Sections 5.1-5.7 - Integrated Review - Operations on Polynomials and Factoring Strategies - Page 313: 42

$20(x+y)(x^2-xy+y^2)$

Factoring the $GCF= 20 $ results to $ 20(x^3+y^3) $. Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of 2 cubes, then, \begin{array}{l} 20(x^3+y^3) \\= 20[(x)+(y)][(x)^2-(x)(y)+(y)^2] \\= 20(x+y)(x^2-xy+y^2) .\end{array}