6.247. Determine the values of $m$ for which one of the roots of the equation $z^{3}-\left(m^{2}-m+7\right) z-\left(3 m^{2}-3 m-6\right)=0$ is -1. Find the other two roots of the equation for these values of $m$.
Solution.

Let $z_{1}=-1$, then $(-1)^{3}-\left(m^{2}-m+7\right)(-1)-\left(3 m^{2}-3 m-6\right)=0$, $m^{2}-m-6=0$, from which $m_{1}=-2, m_{2}=3$. For $m_{1}=-2$ and $m_{2}=3$, the equation becomes $z^{3}-13 z-12=0$. Dividing the left side of the equation by $z+1$: $\frac{z^{3}-13 z-12}{z+1}=z^{2}-z-12, z^{2}-z-12=0$, from which $z_{2}=-3, z_{3}=4$.

Answer: $m_{1}=-2$ and $m_{2}=3$; for these values of $m$, $z_{1}=-1$, $z_{2}=-3, z_{3}=4$.