Task 14. Determine the necessary and sufficient conditions that the numbers $a, b, c$ must satisfy so that the equation $x^{3}+a x^{2}+b x+c=$ $=0(1)$. has three real roots forming an arithmetic progression.
Solution. Let $x=p+t$. After this, the given equation (1) is transformed into:

$$
\begin{gathered}
t^{3}+(a+3 p) t^{2}+\left(3 p^{2}+2 a p+b\right) t+ \\
+\left(p^{3}+a p^{2}+b p+c\right)=0
\end{gathered}
$$

If $p=-a: 3$, i.e., $a+3 p=0$, then the equation takes the form

$$
t^{3}+\left(b-\frac{a^{3}}{3}\right) t+\left(\frac{2 a^{3}}{27}-\frac{a b}{3}+c\right)=0
$$

or

$$
t^{3}=\left(\frac{a^{3}}{3}-b\right) t+\left(\frac{a b}{3}-\frac{2 a^{3}}{27}-c\right)
$$

Let's construct the graphs of the functions $f(t)=t^{3}$ and $\varphi(t)=$ $=\left(\frac{a^{3}}{3}-b\right) t+\left(\frac{a b}{3}-\frac{2 a}{27}-c\right)$ (Fig. 1). Now it is clear that the line $\varphi(t)$ intersects the cubic parabola $f(t)$ at points $A$ and $B$, symmetric with respect to the origin, if

$$
\frac{a b}{3}-\frac{2 a^{3}}{27}-c=0 \text { and } \frac{a^{3}}{3}-b \geqslant 0
$$

If $\frac{a^{3}}{3}-b \geqslant 0$ and $\frac{a b}{3}-\frac{2 a^{3}}{27}-c \neq 0$ and the line $\varphi(t)$ intersects the graph of the function $f(t)$, then

![](https://cdn.mathpix.com/cropped/2024_05_21_8debcc8c8285c4d79a3dg-105.jpg?height=451&width=360&top_left_y=2076&top_left_x=354)

Fig. 1 roots of equation (4) are not symmetric with respect to the origin.

Thus, the necessary and sufficient conditions for the roots of the given cubic equation to be real and form an arithmetic progression are the relations (5).