# Problem 4. Option I.

Find the set of values of the parameter $a$, for which the sum of the cubes of the roots of the equation $x^{2}-a x+a+2=0$ is equal to -8.

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# Solution.

1) $x_{1}^{3}+x_{2}^{3}=\left(x_{1}+x_{2}\right)\left(\left(x_{1}+x_{2}\right)^{2}-3 x_{1} x_{2}\right)=a\left(a^{2}-3(a+2)\right)=a^{3}-3 a(a+2)$.
2) $a^{3}-3 a(a+2)=-8 \Leftrightarrow\left[\begin{array}{l}a=-2, \\ a=1, \\ a=4 .\end{array}\right.$
3) $D=a^{2}-4 a-8$.

For $a=-2 \quad D=4+8-8=4>0$, therefore, $a=-2$ is a solution.

For $a=1 \quad D=1-4-8<0$ - no roots; hence, $a=1$ is not a solution.

For $a=4 \quad D=16-16-8<0$, $a=4$ is also not a solution.

Answer: $\{-2\}$.