Bezborodnikov M.F.

What condition must the coefficients $a, b, c$ of the equation $x^{3}+a x^{2}+b x+c$ satisfy so that its three roots form an arithmetic progression?
Let $x_{1} \leq x_{2} \leq x_{3}$ be the roots of our equation. Then $x_{1}+x_{2}+x_{3}=-a$. For $x_{1}, x_{2}, x_{3}$ to form an arithmetic progression, it is necessary and sufficient that $x_{1}+x_{3}=2 x_{2}$. Therefore, $x_{2}=-a / 3, x_{1}+x_{3}=-$ $2 a / 3$.

Next, $b=x_{1} x_{3}+x_{2}\left(x_{1}+x_{3}\right)=x_{1} x_{3}+\frac{2 a^{2}}{9}$, so $x_{1} x_{3}=b-\frac{2 a^{2}}{9} \cdot$ Thus, $c=-x_{1} x_{2} x_{3}=\frac{a}{3}\left(b-\frac{2 a^{2}}{9}\right)$

For the roots to be real, the system $x_{1}+x_{3}=-2 a / 3, x_{1} x_{3}=b-\frac{2 a^{2}}{9}$ must have real solutions, meaning the quadratic $t^{2}+\frac{2 a}{3} t+b-\frac{2 a^{2}}{9}$ must have a non-negative discriminant. Hence, $\left(\frac{a}{3}\right)^{2} \geq b-\frac{2 a^{2}}{9}, \frac{a^{2}}{3} \geq b$.

## Answer

$27 c=3 a b-2 a^{3}, 3 b \leq a^{2}$