$f\left(x + \frac1x\right)= x^3+x^{-3},$ find $f(x)$

Question

$$f\left(x + \frac1x\right)= x^3+x^{-3},$$ 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.

Answer 1

Maybe $$f(x):=x^3-3x$$

Thus $$f(x+x^{-1})=x^3+3x+3x^{-1}+x^{-3}-3x-3x^{-1}=x^3+x^{-3}$$

Answer 2

$$ f\left(x + \frac1x\right)= x^3+x^{-3} $$ $$ \left(x + \frac1x\right)^3 = x^3 + 3x + 3\frac1x+x^{-3} $$ $$ \Rightarrow f(t) = t^3 - 3t $$

Answer 3

$x^3+x^{-3}=(x + \frac{1}{x})^3-3(x + \frac{1}{x})$

set $t=x + \frac{1}{x}, f(t)= t^3-3t$.

Answer 4

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 $$ f(y) = \begin{cases}y^3 - 3y &\text{ when }x\in(-\infty,-2]\cup [2,\infty) \\ \text{anything} &\text{ when }x\in(-2,2) \end{cases} $$

Answer 5

Hint: $$\left(x + \frac1x\right)^3=x^3+3x^2\frac1x+3x\frac{1}{x^2}+\frac{1}{x^3}$$

Answer 6

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 $x+\frac{1}{x}$.

Answer 7

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...