Let $f(x)=x(4x^2-3)(64x^6-96x^4+36x^2-3)$ and $f^{(n)}=f(f(f(\cdots f(x))\cdots)$ (composed with itself $n$ times). Prove that for all positive integers $n$, $f^{(n)}(x)=x$ has $9^n$ distinct real solutions $x$.
I could only manage to prove this for $n=1$. For that i put $2x=t$ and found the roots of $t(t^2-3)(t^6-6t^4+9t^2-3)=t$
$O$ is an obvious root. So, i proceeded to show that $(m-3)(m^3-6m^2+9m-3)=1$ has $4$ distinct positive roots.
Expanding i got $m^4-9m^3+27m^2-30m+8=0$. $2, 4$ are natural to guess and easy to verify and the other two roots by dividing out $(m-2)$ and $(m-4)$.
Now, someone please help me to prove for $n>1$.