Answer
$y(4-y)(16+4y+y^{2})$
or $-y(y-4)(16+4y+y^{2})$
Work Step by Step
Factor out $y,$
$64 y-y^{4}=y(64-y^{3})$
Recognize 64 as a cube, $64=4^{3}$, so the parentheses hold a difference of cubes, for which we use $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$
$=y\cdot(4-y)(4^{2}+4y+y^{2})$
$=y(4-y)(16+4y+y^{2})$
or, factoring -1 out of the first parentheses,
= $-y(y-4)(16+4y+y^{2})$
