Answer
$(x-y),(x+y),(x^2+y^2),(x^4+y^4)$
Work Step by Step
Given: $x^8-y^8$
Take out the common terms as follows:
$=(x^4)^2-(y^4)^2$
or, $=(x^4-y^4)(x^4+y^4)$
Thus, we have $=(x^2-y^2)(x^2+y^2)(x^4+y^4)$
or, $=(x-y)(x+y)(x^2+y^2)(x^4+y^4)$
factors are $(x-y),(x+y),(x^2+y^2),(x^4+y^4)$