1. Determine the values of $k$ for which the equation

$$
x^{3}-\left(k^{2}-k+7\right) x-\left(3 k^{2}-3 k-6\right)=0
$$

has a root -1. For the values of $k$ thus found, solve the equation.
Solution. For $x=-1$ the equation reduces to $k^{2}-k-6=0$, which has roots $k_{1}=-2$ and $k_{2}=-3$. Substituting $k$ with $k_{1}=-2$ or $k_{2}=-3$ yields the same equation

$$
x^{3}-13 x-12=0
$$

This equation has a root $x_{1}=-1$, so dividing it by $x+1$, we get

$$
x^{3}-13 x-12=(x+1)\left(x^{2}-x-12\right)
$$

Solving the equation $x^{2}-x-12=0$ gives the other two solutions of the equation $x^{3}-13 x-12=0$, which are $x_{2}=-3$ and $x_{3}=4$.