Find all reals $x$,$y$ satisfying the following equation: 
Find all positive reals $x,y \in \mathbb{R}^+$ satisfying:
  $$\frac{x^9}{y} + \frac{y^9}{x} = 10-\frac{8}{xy}$$

Since this involves higher exponents I am unable to tackle this problem. Please help me.
 A: Hint: Write it as:
$x^{10}+y^{10}=10xy-8$
Then use AM-GM:
$10xy-5=x^{10}+y^{10}+1+1+1\ge5\sqrt[5]{x^{10}y^{10}}=5x^2y^2$
Which gives $5(xy-1)^2\le0$, so $xy=1$
Then initial equation can be rewritten as $x^{10}+\frac{1}{x^{10}}=2$
Which by AM-GM again, (or by writing it as a $(x^{10}-1)^2=0$) has the solution, $x^{10}=1$, so $x=\pm1$
So $x=1,y=1$, or $x=-1,y=-1$
A: Just to give a non-AM-GM answer, let $u=xy$, which must be positive in order for $x^{10}+y^{10}=10xy-8$ to have a solution, and note that $x^{10}+y^{10}=(x^5-y^5)^2+2(xy)^5$.  Thus
$$\begin{align}x^{10}+y^{10}-10xy+8
&=(x^5-y^5)^2+2(u^5-5u+4)\\
&=(x^5-y^5)^2+2(u-1)(u^4+u^3+u^2+u-4)\\
&=(x^5-y^5)^2+2(u-1)^2(u^3+2u^2+3u+4)\\
&\ge0
\end{align}$$
with equality if and only if $x^5=y^5$ and $xy=u=1$ (since $u\gt0$ implies $u^3+2u^2+3u+4\gt0$).  From this we see that $x=y=\pm1$ are the only possibilities.  In particular, $(x,y)=(1,1)$ is the only solution with $x,y\in\mathbb{R}^+$.
Remark:  I showed the factorization of $u^5-5u+4$ in two steps for ease of checking.
