Answer
$2(3y-4)(9y^2+12y+16)$
Work Step by Step
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}
