Precalculus, Chapter 9, 9.5, Section 9.5, Problem 37

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 2/x for x, y for y and 4 for n, such that:
(2/x - y)^4 = 4C0 (2/x)^4+4C1 (2/x)^3*(-y)^1+4C2 (2/x)^2*(-y)^2+4C3 (2/x)^1*(-y)^3 + 4C4 3a*(-y)^4
By definition, nC0 = nCn = 1, hence 4C0 = 4C4 = 1.
By definition nC1 = nC(n-1) = n, hence 4C1 = 4C3 = 4.
By definition nC2 = (n(n-1))/2 , hence 4C2 = 6 .
(2/x - y)^4 = 16/(x^4)- (32y)/(x^3)+ (24y^2)/(x^2)- (8y^3)/x + y^4
Hence, expanding the number using binomial theorem yields the simplified result (2/x - y)^4 = 16/(x^4)- (32y)/(x^3)+ (24y^2)/(x^2)- (8y^3)/x + y^4.