Number of real solutions of $f(f(f(x)))=1$, where $f(x) = x-x^{-1}$ given $f(x) = x-x^{-1}$ ,then number of real solution of $f(f(f(x)))=1$ 
$f(x)=\frac{x^2-1}{x}$, $f(f(x))=\frac{(f(x))^2-1}{f(x)} = \frac{(x^2-1)^2-x^2}{x(x^2-1)}=\frac{x^4-3x^2+1}{x^3-x}$
$f(f(f(x))) = \frac{(f(x))^4-3(f(x))^2+1}{(f(x))^3-f(x)} = \frac{(x^2-1)^4-3x^2(x^2-1)^2+x^4}{x[(x^2-1)^3-x^2(x^2-1)]}=1$
how can i solve 8 degree equation

Let $m-\frac{1}{m}=1$, $t-\frac{1}{t}=m$. 
Thus, $x-\frac{1}{x}=t,$ which gives
$$m=\frac{1\pm\sqrt5}{2},$$
$$t^2-\frac{1\pm\sqrt5}{2}t-1=0$$ and
$$x-\frac{1}{x}=\frac{1\pm\sqrt5\pm\sqrt{22\pm2\sqrt5}}{4}.$$
From the last inequality we can get all $8$ distinct real roots.
