= Solution
The algebraic identity
$$
x^3+x^{-3}=(x+x^{-1})^3-3(x+x^{-1})
$$
suggests \b[$f(t)=t^3-3t$ for every $t>0$], and substitution verifies the <functional equation>.
In fact $x+1/x\geq2$, since $(x-1)^2/x\geq0$. Conversely, for each $t\geq2$, the <quadratic equation> $x^2-tx+1=0$ has a positive root $(t+\sqrt{t^2-4})/2$. Thus the <functional equation> forces $f(t)=t^3-3t$ exactly on $[2,\infty)$; its values on $(0,2)$ are arbitrary. The displayed <function> is one convenient choice on the full required domain.
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