$f\left(x + \frac1x\right)= x^3+x^{-3},$ find $f(x)$ $$f\left(x + \frac1x\right)= x^3+x^{-3},$$ find $f(x)$.
What i do know at this state is that..
express x as a function of y :
$y= x + 1/x$
$x^2−xy+1=0$
Quad formula: 
$x= (y ± \sqrt {y^2-4}) / 2$
When i substitute this into the original equation, i can't solve it.

As $\left|x+  \frac 1x\right| \ge 2$,  $f$ is only constrained of $\{y :|y|\ge 2  \}$.
$$ x + \frac 1x = y 
\\ x^2 - xy + 1  = 0
\\ x = \frac12\left(
y \pm \sqrt{y^2 - 4}
\right)
$$
Hence on this set:
$$f(y) = \frac 18\left(y + \sqrt{y^2 - 4}\right)^3 + 
8\left(y + \sqrt{y^2 - 4}\right)^{-3}
$$
When you expand the formula, it turns out that 
$$
f(y) = \begin{cases}y^3 - 3y &\text{ when }x\in(-\infty,-2]\cup [2,\infty) \\
\text{anything} &\text{ when }x\in(-2,2)
\end{cases}
$$
