Problem 881

Answer: $a= \pm 4$.

Solution. For the equation to have roots, its discriminant must be positive, hence $a^{2}-12>0$. Under this condition, by Vieta's theorem, $x_{1}+x_{2}=-a, x_{1} x_{2}=3$. Then $x_{1}^{2}+$ $x_{1} x_{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}=a^{2}-3$.

Transform the given equality:

$$x_{1}^{3}-x_{2}^{3}-39 \cdot \frac{x_{1}-x_{2}}{x_{1} x_{2}}=0 \Leftrightarrow\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)-\frac{39\left(x_{1}-x_{2}\right)}{x_{1} x_{2}}=0$$

Since the roots are distinct, $x_{1}-x_{2} \neq 0$. Dividing both sides by $x_{1}-x_{2}$ and substituting the values given above, we get $a^{2}-3-\frac{39}{3}=0 \Leftrightarrow a^{2}-16=0$, from which $a= \pm 4$. Both found values of the parameter satisfy the inequality $a^{2}-12>0$.