Answer
$(x^4+1)(x^2+1)(x+1)(x-1)$
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: $x^8-1=\\=x^8-1^8\\=(x^4)^2-(1^4)^2\\=(x^4+1^4)(x^4-1^4)\\=(x^4+1)((x^2)^2-(1^2)^2)\\=(x^4+1)(x^2+1^2)(x^2-1^2)\\=(x^4+1)(x^2+1)(x+1)(x-1)$
