Answer
$ x^{10}-5x^8+10x^6-10x^4+5x^2-1$
Work Step by Step
We will apply the Binomial Theorem or Binomial expansion as follows:
$(x+y)^n= \dbinom{n}{0}x^ny^0+ \dbinom{n}{1}x^{n-1}y^1+........+ \dbinom{n}{n}x^0y^n $
$(x^2-1)^5=x^{10}-\dfrac{5!}{4! 1!}x^8+\dfrac{5!}{3! 1!}x^6-\dfrac{5!}{2! 3!}x^4+\dfrac{5!}{1! 4!}x^2-1$
$=x^{10}-5x^8+10x^6-10x^4+5x^2-1$
