## Intermediate Algebra (6th Edition)

$a^2(2b+3a)(4b^2-6ab+9a^2)$
Factoring the $GCF= a$ results to $a(8b^3+27a^3)$. Using $a^3+b^3=(a+b)(a^2-ab+b^2)$, or the factoring of the sum/difference of two cubes, then, \begin{array}{l} a(8b^3+27a^3) \\= a^2[(2b)+(3a)][(2b)^2-(2b)(3a)+(3a)^2) \\= a^2(2b+3a)(4b^2-6ab+9a^2) .\end{array}