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.

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 $x=-1,y=-1$
