Х^3-х^2>2х неравенствооо

Решите задачу:

$x^3-x^2\ \textgreater \ 2x\\\\ x^3-x^2-2x\ \textgreater \ 0\\\\ x(x^2-x-2)\ \textgreater \ 0\\\\ x=0\\\\\\ x^2-x-2=0\\\\ x^2-x=2\\\\ x^2- \frac{x}{2}+ \frac{1}{4}=2+ \frac{1}{4}\\\\ (x- \frac{1}{2})^2= \frac{9}{4}\\\\ \sqrt{(x- \frac{1}{2})^2} = \sqrt{ \frac{9}{4}}\\\\ |x- \frac{1}{2}|=\pm \frac{3}{2}\\\\ x- \frac{1}{2}=- \frac{3}{2}\\\\ x=- \frac{3}{2}+ \frac{1}{2}\\\\ x_1=-1\\\\\\ x- \frac{1}{2}= \frac{3}{2}\\\\ x= \frac{3}{2}+ \frac{1}{2}\\\\ x_2=2\\\\ x\in(-1;0)\bigcup(2;+\infty)$