<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] I first would eliminate the negative exponents of $x$ by multiplying $f(x)$ by $x^3$ to get
$$g(x)=x^3f(x)=x^6-12\,x^4+48\,x^2-64$$
If you are right in assuming that $f(x)=h(x)^3$ then $g(x)=p(x)^3$ and $g'(x)=3p'(x)p(x)^2$
Now we assume that $p$ is a polynomial then
$$p(x)^2 \mid \text{gcd}(g(x),g'(x))$$
and 
$$\frac{g(x)}{\text{gcd}(p(x),p'(x))} \mid p(x)$$
We get
$$p'(x)=6\,x^5-48\,x^3+96\,x$$
and 
$$\text{gcd}(p(x),p'(x))=x^4-8\,x^2+16$$
$$\frac{g(x)}{\text{gcd}(p(x),p'(x))}=x^2-4=(x-2)(x+2)$$
Because the degree of the last polynomial is $2$ we must have
$$p(x)=x^2-4$$
(ignoring constant factors)
We can check that $$x^6-12\,x^4+48\,x^2-64=(x^2-4)^3=(x-2)^3(x+2)^3$$
and so
$$f(x)=(x-\frac{4}{x})^3=(\sqrt{x}-\frac{2}{\sqrt{x}})^3 (\sqrt{x}+\frac{2}{\sqrt{x}})^3$$
Now let's solve
$$(x-2/x)=y$$
for $x$. By multiplying the equation by $x<$ we finally get the quadratic equation
$$x^2-yx-2=0$$
and get
$$x=-{{\sqrt{y^2+8}-y}\over{2}} $$
$$x=+{{\sqrt{y^2+8}+y}\over{2}}$$
So we have two inverse images for each value $y$.
</s>