Equivalent function inputs

Question

$$f\left(x-\frac 1x\right) = x^3-\frac 1{x^3}\implies f(-x) = \,?$$

I can't think of anything considering that the input of the function that we already know its definition is not $x$ but a composite.

Answer 1

Write $t=x-1/x$ then $$ t^3 =x^3-3x+3/x-1/x^3 = f(t)-3t$$ so $$f(t) = t^3+3t\implies f(-t)=-t^3+3t$$

Notice that $x\mapsto x-1/x$ is also surjective from $\mathbb{R}\setminus\{0\} \to \mathbb{R}$.

Answer 2

After

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

we have

$$ x-\frac{1}{x} = \frac{1}{3}\left(f\left(x-\frac{1}{x}\right)-\left(x-\frac{1}{x}\right)^3\right) $$

or

$$ y = \frac{1}{3}\left(f(y)-y^3\right) $$

then

$$ f(-x) = -(3+x^2)x $$

Answer 3

Use: $$a^3-b^3=(a-b)(a^2+ab+b^2)=(a-b)((a-b)^2+3ab).$$

Note that: $$\begin{align}f\left(x-\frac 1x\right) &= x^3-\frac 1{x^3}=\left(x-\frac1x\right)\left(\left(x-\frac1x\right)^2+3\right) \Rightarrow \\ f(x)&=x(x^2+3) \Rightarrow \\ f(-x)&=-x(x^2+3).\end{align}$$