Answer
$(y-4)(y^2+4y+16)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
y^3-64
,$ use the factoring of the sum/difference of $2$ cubes.
$\bf{\text{Solution Details:}}$
Using $(a\pm b)(a^2\mp ab+b^2)$ or the factoring of the sum/difference of $2$ cubes, the expression above is equivalent to
\begin{array}{l}\require{cancel}
(y-4)[(y)^2+y(4)+(4)^2]
\\\\=
(y-4)(y^2+4y+16)
.\end{array}