Chapter 5 - Sections 5.1-5.7 - Integrated Review - Operations on Polynomials and Factoring Strategies - Page 313: 26

$27x^2y(xy-2)(x^2y^2+2xy+4)$

Factoring the $GCF= 27x^2y $ results to $ 27x^2y(x^3y^3-8) $. Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of two cubes, then, \begin{array}{l} 27x^2y(x^3y^3-8) \\= 27x^2y[(xy)+(-2)][(xy)^2-(xy)(-2)+(-2)^2] \\= 27x^2y(xy-2)(x^2y^2+2xy+4) .\end{array}