Chapter 5 - Sections 5.1-5.7 - Integrated Review - Operations on Polynomials and Factoring Strategies - Page 313: 45

$a^3b(4b-3)(16b^2+12b+9)$

Factoring the $GCF=a^3b$ results to $a^3b(64b^3-27)$. Using $a^3+b^3=(a+b)(a^2-ab+b^2)$ or the factoring of two cubes, then, \begin{array}{l} a^3b(64b^3-27) \\= a^3b[(4b)+(-3)][(4b)^2-(4b)(-3)+(-3)^2] \\= a^3b(4b-3)(16b^2+12b+9) .\end{array}