 Solution

Put $a=2/\sqrt{3}$. On $[-a,a]$, the extrema of $|x(1-x^2)|$ occur at $\pm 1/\sqrt{3}$ or the endpoints. The corresponding values are $2/(3\sqrt{3})$, including the endpoint value $a(a^2-1)$. ConsequentlyThus $[-a,a]$ is a trapping region andIt contains a neighbourhood of the origin. Sign changes are possible because $f(x)$ has opposite sign to $x$ for $1<|x|<a$.