Answer
$x^8-4x^6+6x^4-4x^2+1$
Work Step by Step
Apply Binomial Theorem.
$(a+b)^n=\sum_{r=0}^n\dbinom{n}{r}(a)^{n-r}b^r$
$(x^2-1)^4=\dbinom{4}{0}(x^2)^4(-1)^0+\dbinom{4}{1}(x^2)^3(-1)^1+\dbinom{4}{2}(x^2)^2(-1)^2+\dbinom{4}{3}(x^2)^1(-1)^3+\dbinom{4}{4}(x^2)^0(-1)^4$
or, $=x^8+(4)(x^6)(-1)+6x^4+4x^2(-1)+1$
or, $=x^8-4x^6+6x^4-4x^2+1$
