Answer
$[(y-5)^2+z^4](y-5+z^2)(y-5-z^2)$
Work Step by Step
Using $x^2-y^2=(x+y)(x-y)$ or the factoring of the difference of 2 squares, then the factored form of the given expression, $
(y-5)^4-z^8
,$ is
\begin{array}{l}
[(y-5)^2+z^4][(y-5)^2-z^4]
\\\\=
[(y-5)^2+z^4][(y-5)+z^2][(y-5)-z^2]
\\\\=
[(y-5)^2+z^4](y-5+z^2)(y-5-z^2)
\end{array}