If $x+\frac {1}{x}=2,x\neq 0$$x^{99}+\frac {1}{x^{99}}=?$

We have, $x^2+1-2x=0$

$\Rightarrow (x-1)^2=0$

$\Rightarrow x=1$

$\therefore x^{99}+\frac {1}{x^{99}}=1+1=2$