Proving there are no periodic points I have the following function

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

By inspection, I found that the only periodic points are $0, \pm \sqrt{2}$ with period $1$ (i.e. fixed points).

Nevertheless, I don't know how to prove that there's no periodic points other than the fixed points I found.
Let $f(x)=x^3-x$

$a^3-a=b^3-b$ where $a\ne b$

$\implies0=a^3-b^3-(a-b)=(a-b)(a^2+ab+b^2-1)$

So, we need $a^2+ab+b^2-1=0\implies a=\dfrac{-b\pm\sqrt{b^2-4(b^2-1)}}2=\dfrac{-b\pm\sqrt{4-3b^2}}2$

For real $a,$ we need $4-3b^2\ge0\iff3b^2\le4$

In that case $f(a)=f(b)$ right?
Let $f(x)=x^3-x$

$a^3-a=b^3-b$ where $a\ne b$

$\implies0=a^3-b^3-(a-b)=(a-b)(a^2+ab+b^2-1)$

So, we need $a^2+ab+b^2-1=0\implies a=\dfrac{-b\pm\sqrt{b^2-4(b^2-1)}}2=\dfrac{-b\pm\sqrt{4-3b^2}}2$

For real $a,$ we need $4-3b^2\ge0\iff3b^2\le4$

In that case $f(a)=f(b)$ right?