<s>[INST] Given the function $f(x)=\frac {1}{\sqrt[3] {1-x^3}}$ find $\underbrace {f(f(\cdots f(19)\cdots))}_{95}$ Given the function $$f(x)=\frac {1}{\sqrt[3] {1-x^3}}$$ find $$\underbrace {f(f(\cdots f(19)\cdots))}_{95}$$
My try:
Define $$f^n(x)=\underbrace {f(f(\cdots f(x)\cdots))}_{n}$$
We see that  $$f(x)=\frac {1}{\sqrt[3] {1-x^3}}$$
$$f^2(x)=\frac {\sqrt[3] {1-x^3}}{-x}$$
$$f^3(x)=x$$
And again $$f^4(x)=\frac {1}{\sqrt[3] {1-x^3}}$$
Hence $$f^{3k+2}(x)=\frac {\sqrt[3] {1-x^3}}{-x}$$
Substituting $k=31$ and $x=19$ I can find $$\underbrace {f(f(\cdots f(19)\cdots))}_{95}$$
So am I going on the right path. 
 [/INST] As you have already said there are only 3 interactions, then it goes back to the original function:
\begin{align}
f_0 = \underbrace{f(x)}_1&=\frac{1}{\sqrt[3] {1-x^3}} = \left(\frac{1}{1-x^3}\right)^{\frac{1}{3}}\\
f_1 = \underbrace{f(f(x))}_2&= \left(\frac{1}{1-\left(\frac{1}{1-x^3}\right)}\right)^{\frac{1}{3}} = \left(1 - \frac1{x^3}\right)^{\frac{1}{3}}\\
f_2 =\underbrace{f(f(f(x)))}_3 &= \left(\frac{1}{1-\left(1 - \frac1{x^3}\right)}\right)^{\frac{1}{3}} = x  
\end{align}
If you count from $0$, so it's the $95$-th interaction equals $f_{94}$. It follows $94 \pmod 3 = 1$,which implies $f_{94} = f_1$:
$$\underbrace{f(f(\cdots f(19)\cdots))}_{95} = f_{94} = f_1 =\left(1 - \frac1{19^3}\right)^{\frac{1}{3}} $$
</s>