Answer
$x^{10}-5x^8+10x^6-10x^4+5x^2-1$
Work Step by Step
Apply Binomial Theorem.
$(a+b)^n=\sum_{r=0}^n\dbinom{n}{r}(a)^{n-r}b^r$
$(x^2-1)^5=\dbinom{5}{0}(x^2)^5(-1)^0+\dbinom{5}{1}(x^2)^4(-1)^1+\dbinom{5}{2}(x^2)^3(-1)^2+\dbinom{5}{3}(x^2)^2(-1)^3+\dbinom{5}{4}(x^2)^1(-1)^4+\dbinom{5}{5}(x^2)^0(-1)^5$
or, $=x^{10}-5x^8+10x^6+10x^4(-1)+5x^2+(1)(1)(-1)$
or, $=x^{10}-5x^8+10x^6-10x^4+5x^2-1$
