Answer
$(x^4+y^4)(x^2+y^2)(x+y)(x-y)$
Work Step by Step
The product of a binomial sum and a binomial difference is: $(A+B)(A-B)=A^2-B^2$.
The square of a binomial sum is: $(A+B)^2=A^2+2AB+B^2$.
The square of a binomial difference is: $(A-B)^2=A^2-2AB+B^2$.
Hence here: $x^8-y^8=\\=(x^4)^2-(y^4)^2\\=(x^4+y^4)(x^4-y^4)\\=(x^4+y^4)((x^2)^2-(y^2)^2)\\=(x^4+y^4)(x^2+y^2)(x^2-y^2)\\=(x^4+y^4)(x^2+y^2)(x+y)(x-y)$
