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.
 A: Intuitively, one should let $a=x+y$ and $b=x-y$.
Then from the condition $\sqrt {xy}=1$ one obtains $a^2-b^2=4$.
The other condition becomes
$$a = \sqrt[3]{\frac ab} + \sqrt[3]{\frac ba}$$
Cubing both sides, we have
\begin{align}a^3 &= \frac ab + 3\left(\sqrt[3]{\frac ab}\right)^2\sqrt[3]{\frac ba}+ 3\sqrt[3]{\frac ab}\left(\sqrt[3]{\frac ba}\right)^2+ \frac ba
\\&=\frac ab + \frac ba+ 3\left(\sqrt[3]{\frac ab} + \sqrt[3]{\frac ba}\right)
\\&=a\left(\frac 1b + \frac b{a^2}\right)+ 3a\end{align}
Eliminating $a$ from both sides gives
$$a^2 = \frac 1b + \frac b{a^2} + 3$$
$$4+b^2=\frac1b + \frac b{4+b^2} + 3$$
$$1+b^2=\frac {4+b^2+b^2}{b(4+b^2)}$$
$$b(4+b^2)(1+b^2)-2b^2 =4$$
Expanding $b(4+b^2)(1+b^2)-2b^2$ gives $b^5+5b^3-2b^2+4b$, which seems very familiar...
A: 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
$$ \boxed{v^5 + 5v^3 -2v^2 + 4v =4}$$
