Precalculus, Chapter 9, 9.5, Section 9.5, Problem 33

You need to use the binomial formula, such that:
(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k) y^k
You need to replace x^2 for x, y^2 for y and 4 for n, such that:
(x^2+y^2)^4 = 4C0 (x^2)^4 + 4C1 (x^2)^3*(y^2)^1 + 4C2 (x^2)^2*(y^2)^2 +4C3 (x^2)*(y^2)^3 +4C4 (y^2)^4
By definition, nC0 = nCn = 1, hence 4C0 = 4C4 = 1 .
By definition nC1 = nC(n-1) = n, hence 4C1 = 4C3 = 4.
By definition nC2 = nC(n-2) = (n(n-1))/2 , hence 4C2 = (4(4-1))/2 = 6
(x^2+y^2)^4 = x^8 + 4(x^6)*(y^2) +6 (x^4)*(y^4) +4 (x^2)*(y^6)+ (y^8)
Hence, expanding the complex number using binomial theorem yields the simplified result (x^2+y^2)^4 = x^8 + 4(x^6)*(y^2) +6 (x^4)*(y^4) +4 (x^2)*(y^6)+ (y^8).