Problem 9.1. Find all real numbers $a$ such that the roots $x_{1}$ and $x_{2}$ of the equation

$$
x^{2}+6 x+6 a-a^{2}=0
$$

satisfy the relation $x_{2}=x_{1}^{3}-8 x_{1}$.

Ivan Landjev
9.1. It follows from Vieta's formulae that

$$
-6=x_{1}+x_{2}=x_{1}+x_{1}^{3}-8 x_{1}
$$

Therefore $x_{1}^{3}-7 x_{1}+6=0$ and $x_{1}=-3,1$ or 2 . Plugging these values of $x_{1}$ in the initial equation gives $a=3$ for $x=-3, a=-1$ and 7 for $x=1$ and $a=-2$ and 8 for $x=2$.