<s>[INST] How to find the $f^{-1}(x)$ of $f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}}$ It is a question from a quiz.
The following is the whole question.

Let
  \begin{eqnarray}
\\f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}} , \space x\in (-\infty ,0),
\end{eqnarray}
  find $f^{-1}(x)$. Hint : $f(x)$ can be written in the form, like$(A+B)^{3}$.

The first thing I think is $(A+B)^{3}=A^3+3A^2B+3AB^2+B^3$, then try to make it become the the form of $A^3+3A^3B+3AB^3+B^3$. However, it it so difficult to obtain this form. 
I need help.
Update :
Now I have $\left(x - \frac 4x\right)^3$ but how to find the $f^{-1}(x)$ of $f(x)=\left(x - \frac 4x\right)^3$?

Thank you for your attention
 [/INST] for $x \in (-\infty,0)$  the  inverse function has the form 
$$
f^{-1}(x)=\frac{1}{2}\left( \sqrt[3]{x}-\sqrt{\sqrt[3]{x^2}+16} \right)
$$
As mentioned above  we have 
$y = \left(x - \dfrac4x\right)^3 \Longrightarrow x^2 - x\sqrt[3]{y} - 4 = 0$.
By solving the equation we  get 
$$
x=\frac{1}{2}\left( \sqrt[3]{y}-\sqrt{\sqrt[3]{y^2}+16} \right),
$$
and 
$$
x=\frac{1}{2}\left( \sqrt[3]{y}+\sqrt{\sqrt[3]{y^2}+16} \right).
$$
The first expression is always negative (note $x \in (-\infty,0)$ but second is  positive and  not suitable for us. Thus, changing the variables, we  get the inverse function.
</s>