<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] First of all, $f$ does not have a inverse function because you always have two root of $f - y$. One positive root, one negative root.
If you add the constraints that $x$ is positive and $y$ is real, then there will be an inverse function.
$$(x - \frac4x)^3=y$$
Since $y$ is real:
$$x - \frac4x=\sqrt[3]y$$
Since $x$ is not zero.
$$x^2 - 4=\sqrt[3]yx$$
Solve the quadratic equation:
$$x^2 - \sqrt[3]yx = 4 $$
$$(x-\frac{\sqrt[3]y}2)^2=4+(\frac{\sqrt[3]y}2)^2$$
$$ x - \frac{\sqrt[3]y}2 = \pm \sqrt{4+(\frac{\sqrt[3]y}2)^2}$$
$$ x = \frac{\sqrt[3]y}2 \pm \sqrt{4+(\frac{\sqrt[3]y}2)^2}$$
Since $x$ is positive, and the magnitude of the second term is greater than the first term, so you have to take the plus sign. This is your inverse function:
$$f^{-1}(y) = \frac{\sqrt[3]y}2 + \sqrt{4+(\frac{\sqrt[3]y}2)^2}$$
If, however, you constrain $x$ to be negative, then take the minus sign above.
</s>