Решить неравенства методом интервалов(x + 1)^3(-2x + 6)/3x+21<0

$\frac{(x + 1)^3(-2x + 6)}{3x+21}\ \textless \ 0\\\\ \frac{(x+1)^3*(-2)*(x-3)}{3*(x+7)}\ \textless \ 0\ \ |\ \ *\ \ -\frac{3}{2}\\\\ \frac{(x+1)^3*(x-3)}{x+7}\ \textgreater \ 0\\\\ \frac{[x-(-1)]^3*[x-3]}{x-(-7)}\ \textgreater \ 0\\\\ ------(-7)+++++(-1)-----(3)+++++\ \textgreater \ x\\\\ x\in(-7;\ -1)\cup(3;\ +\infty)$

Ответ: $(-7;\ -1)\cup(3;\ +\infty)$