Find alternate form of equation involving roots

Question

I am trying to find out why

$$(x-y)^{3/2}*x^{-3/2} = (1-\frac{y}{x})^{3/2}$$

But I can't for the life of me find a way of deriving it although it looks painfully simple. What am I missing?

Answer 1

You have: $$(x-y)^\frac{3}{2}=\left( x (1-y/x) \right)^\frac{3}{2}=x^\frac{3}{2}(1-y/x)^\frac{3}{2}$$ and then $x^\frac{3}{2} \times x^\frac{-3}{2}=1$.

Answer 2

Hint $$(x-y)^{3/2}*x^{-3/2} = (1-\frac{y}{x})^{3/2}$$ $$(x-y)^{3/2}*\frac 1 {x^{3/2}} = (1-\frac{y}{x})^{3/2}$$ $$(\frac {x-y}x)^{3/2} = (1-\frac{y}{x})^{3/2}$$ Therefore $$(1-\frac {y}x)^{3/2} = (1-\frac{y}{x})^{3/2}$$

Answer 3

$$ (x-y)^{3/2}x^{-3/2} =\left ( (x-y)^3x^{-3}\right )^{1/2}\\= \left ( (x-y)^3\frac{1}{x^3}\right )^{1/2} \\ =\left ( \frac{x^3-3x^2y+3xy^2-y^3}{x^3}\right )^{1/2}\\ = \left( 1-3\frac y x +3\frac{y^2}{x^2} -\frac{y^3}{x^3}\right )^{1/2}\\ =\left ( (1- \frac{y}{x})^3 \right )^{1/2} \\= \left ( 1- \frac{y}{x} \right )^{3/2} $$