Answer
$2(y-2)(y^{2}+2y+4)$
Work Step by Step
$ 2y^{3}-16\qquad$...factor out the common term, $2$.
$=2(y^{3}-8)\qquad$...use the difference of two cubes:
$\mathrm{A}^{3}-\mathrm{B}^{3} = (\mathrm{A}-\mathrm{B})(\mathrm{A}^{2} + \mathrm{A}\mathrm{B} + B^{2})$
$=2((y)^{3}-(2)^{3})$
$=2(y-2)(y^{2}+2y+4)$
