$f\left(x + \frac1x\right)= x^3+x^{-3},$ 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.
 A: There is a pretty straightforward answer. Polynomials are very strict in what terms they allow. Why not try matching the first part $x^3$? What would you have to do with the term $x+\frac{1}{x}$ to get $x^3$ as a first term? Then look at your result and try to remove whatever there is too much, again in terms of $x+\frac{1}{x}$.
A: $x^3+x^{-3}=(x + \frac{1}{x})^3-3(x + \frac{1}{x})$
set $t=x + \frac{1}{x}, f(t)= t^3-3t$.
A: $$
f\left(x + \frac1x\right)= x^3+x^{-3}
$$
$$
\left(x + \frac1x\right)^3 = x^3 + 3x + 3\frac1x+x^{-3}
$$
$$
\Rightarrow f(t) = t^3 - 3t
$$
A: 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 
$$
f(y) = \begin{cases}y^3 - 3y &\text{ when }x\in(-\infty,-2]\cup [2,\infty) \\
\text{anything} &\text{ when }x\in(-2,2)
\end{cases}
$$
A: Maybe
$$f(x):=x^3-3x$$
Thus
$$f(x+x^{-1})=x^3+3x+3x^{-1}+x^{-3}-3x-3x^{-1}=x^3+x^{-3}$$
A: Hint: $$\left(x + \frac1x\right)^3=x^3+3x^2\frac1x+3x\frac{1}{x^2}+\frac{1}{x^3}$$
A: Hmm, $x+1/x = 2 \cosh( \log (x))$ so what you have is also $$f(2 \cosh(\log(x))) = 2\cosh(3\log(x)) $$ 
Perhaps you can then continue on your own...
