Suppose $x$ and $y$ are unequal real numbers 
Suppose $x$ and $y$ are unequal real numbers. If
$$
\sqrt[3]{\frac{x+y}{x-y}} + \sqrt[3]{\frac{x-y}{x+y}} = x+y
\qquad \text{and} \qquad
\sqrt{xy}=1
$$
then find the value of
$$
(x-y)^5 + 5(x-y)^3 - 2(x-y)^2+ 4(x-y)
$$

For the above question, I got $x$ not equal to $-1,0,1$ and $y = 1/x$.
Then, I tried finding $x-y$ using $\big((x+y)^2-4xy)\big)^{\frac{1}{2}}$ and then solving for $x$ and $y$.
I think I over complicated it and there must be some easier way. Please help me out.

Tricky one.
Let $x+y = u$, $x-y=v$.
$$ xy=1 \Rightarrow u^2-v^2 =4$$
and cubing original equation :$$ \sqrt[3]{\dfrac{u}{v}} + \sqrt[3]{\dfrac{v}{u}} = u $$
gives
$$ \dfrac{u}{v} + \dfrac{v}{u} + 3u = u^3 $$
$$ \Rightarrow u^2 + v^2 + 3u^2v = u^4v$$
Substitute $u^2=4+v^2$ ,
$$(4+v^2) + v^2 + 3(4+v^2)v = (4+v^2)^2v $$
directly yields
$$ \boxed{v^5 + 5v^3 -2v^2 + 4v =4}$$
