Answer
$(x^2-y^2)^6=(x^2+(-y^2))^6=\binom{6}{0}(x^2)^6(-y^2)^0+\binom{6}{1}(x^2)^5(-y^2)^1+\binom{6}{2}(x^2)^4(-y^2)^2+\binom{6}{3}(x^2)^3(-y^2)^3+\binom{6}{4}(x^2)^2(-y^2)^4+\binom{6}{5}(x^2)^1(-y^2)^5+\binom{6}{6}(x^2)^0(-y^2)^6=x^{12}-6x^{10}y^2+15x^8y^4-20x^6y^6+15x^4y^8-6x^2y^{10}+y^{12}$
Work Step by Step
The binomial theorem expands an algebraic expression in the form of:
$(x+y)^n=\binom{n}{0}x^ny^0+\binom{n}{1}x^{n-1}y^1+\binom{n}{2}x^{n-2}y^2+...+\binom{n}{i}x^{n-i}y^i+...+\binom{n}{n-1}x^{1}y^{n-1}+\binom{n}{n}x^0y^n=\sum_{i=0}^{n}\binom{n}{i}x^{n-i}y^i$
Here, x becomes $x^2$ and y becomes $-y^2$, $n=6$
$(x^2-y^2)^6=(x^2+(-y^2))^6=\binom{6}{0}(x^2)^6(-y^2)^0+\binom{6}{1}(x^2)^5(-y^2)^1+\binom{6}{2}(x^2)^4(-y^2)^2+\binom{6}{3}(x^2)^3(-y^2)^3+\binom{6}{4}(x^2)^2(-y^2)^4+\binom{6}{5}(x^2)^1(-y^2)^5+\binom{6}{6}(x^2)^0(-y^2)^6=x^{12}-6x^{10}y^2+15x^8y^4-20x^6y^6+15x^4y^8-6x^2y^{10}+y^{12}$
