Precalculus, Chapter 9, 9.5, Section 9.5, Problem 25

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 for x, y for y and 5 for n, such that:
(x+y)^5 = 5C0 (x)^5+5C1 (x)^4*(y)^1+5C2 (x)^3*(y)^2+5C3 x^2*y^3 + 5C4 x*y^4 + 5C5 y^5
By definition, nC0 = nCn = 1, hence 5C0 = 5C5 = 1.
By definition nC1 = nC(n-1) = n, hence 5C1 = 5C4 = 5.
By definition nC2 = (n(n-1))/2 , hence 5C2 = 5C3 = 10.
(x+y)^5 = x^5+5x^4*y+10x^3*y^2+10x^2*y^3 +5x*y^4+y^5
Hence, expanding the complex number using binomial theorem yields the simplified result (x+y)^5 = x^5+5x^4*y+10x^3*y^2+10x^2*y^3 +5x*y^4+y^5.