Intermediate Algebra (6th Edition)

$2(3y-4)(9y^2+12y+16)$
Factoring the $GCF= 2$ results to $2(27y^3-64)$. 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} 2(27y^3-64) \\= 2[(3y)+(-4)][(3y)^2-(3y)(-4)+(-4)^2) \\= 2(3y-4)(9y^2+12y+16) .\end{array}