## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$[(y-5)^2+z^4](y-5+z^2)(y-5-z^2)$
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}