Precalculus, Chapter 9, 9.5, Section 9.5, Problem 38

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, 3y for y and 5 for n, such that:
(2/x-3y)^5 = 5C0 (2/x)^5 +5C1 (2/x)^4*(-3y)^1+5C2 (2/x)^3*(-3y)^2 + 5C3 (2/x)^2 (-3y)^3 + 5C4 (2/x)(-3y)^4 + 5C5 (-3y)^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.
(2/x-3y)^5 = 32/x^5 - (240y)/x^4 + (720y^2)/x^3 - (1080y^3)/x^2 + (810y^4)/x - 243 y^5
Hence, expanding the expression using binomial theorem yields (2/x-3y)^5 = 32/x^5 - (240y)/x^4 + (720y^2)/x^3 - (1080y^3)/x^2 + (810y^4)/x - 243 y^5.