By rules of the domain ofthe function that denominator can not be zero, we have

\(\displaystyle{x}^{{3}}-{x}\frac{=}{{0}}\)

\(\displaystyle{x}^{{3}}-{x}\) can be written as \(\displaystyle{x}^{{3}}-{x}={x}{\left({x}^{{2}}-{1}\right)}\)

From here we have that

\(\displaystyle{x}^{{3}}—{x}={x}{\left({x}^{{2}}-{1}\right)}={0}\) when \(x = 0\) or \(\displaystyle{x}^{{2}}—{1}={0}\).

\(\displaystyle{x}^{{2}}—{1}={0}\) is quadratic function and we have

\(\displaystyle{x}^{{2}}={1}\)

Then domain of the given function is

\(R\{-1,0,1\}\)