Problem 4. II variant.

Find the set of values of the parameter $a$, for which the sum of the cubes of the roots of the equation $x^{2}+a x+a+1=0$ is equal to 1.
# Solution.

1) $x_{1}^{3}+x_{2}^{3}=\left(x_{1}+x_{2}\right)\left(\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right)=-a\left(a^{2}-3(a+1)\right)=-a^{3}+3 a(a+1)$.
2) $-a^{3}+3 a(a+1)=1 \Leftrightarrow\left[\begin{array}{l}a=-1, \\ a=2 \pm \sqrt{3} \text {. }\end{array}\right.$
3) $D=a^{2}-4 a-4$.

For $a=-1 \quad D=1+4-4>0$, therefore, $a=-1$ is a solution.

For $a=2 \pm \sqrt{3} \quad D=7 \pm 4 \sqrt{3}-8 \mp 4 \sqrt{3}-4=-5<0$; hence, the values $a=2 \pm \sqrt{3}$ are not solutions.

Answer: $\{-1\}$.