Need factoring this binomial: x^4-1?

Use the difference of squares factorization - that for any numbers a and b, (a-b)(a+b)=a^2-b^2.

We have:

(x^2+1)(x^2-1)=x^4-1

In addition:

(x-1)(x+1)=x^2-1, so we have:

(x^2+1)(x+1)(x-1)

As our complete factorization.

So we notice that is the special factorization known as 'the difference of two perfect squares'
exg
a^2-b^2=(a-b)(a+b) so
x^4=(x^2)^2 and 1=1^2 so
(x^2)^2-(1)^2=(x^2-1)(x^2+1)

Is there more to the question?