Example 1. The roots of the equation

$$
x^{3}+a x^{2}+b x+c=0
$$

form a geometric progression. What necessary and sufficient condition must the coefficients of the equation satisfy?
Let $x_{1}, x_{2}, x_{3}$ be the roots of the equation. From the condition, it follows that

$$
x_{2}^{2}=x_{1} x_{3}
$$

Multiply both sides of this equation by $x_{2}: x_{2}^{3}=x_{1} x_{2} x_{3}$. Since $x_{1} x_{2} x_{3}=-c$, then

$$
x_{2}^{3}=-c
$$

Moreover, we have:

$$
\begin{aligned}
& x_{1}+x_{2}+x_{3}=-a, \\
& x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3}=b
\end{aligned}
$$

Considering equation (2), we get:

$$
x_{1} x_{2}+x_{2} x_{3}+x_{2}^{2}=b, \text{ or } x_{2}(-a)=b .
$$

From equations (3) and (4), the dependence between the coefficients follows:

$$
a^{3} c=b^{3}
$$

Conversely, let the equation be of the form:

$$
x^{3}+a x^{2}+b x+\frac{b^{3}}{a^{3}}=0
$$

Group the terms of the equation as follows:

$$
\left(x^{3}+\frac{b^{3}}{a^{3}}\right)+a x\left(x+\frac{b}{a}\right)=0
$$

or

$$
\left(x+\frac{b}{a}\right)\left(x^{2}-\left(\frac{b}{a}-a\right) x+\frac{b^{2}}{a^{2}}\right)=0
$$

Hence,

$$
\begin{gathered}
x_{2}=-\frac{b}{a} \quad\left(\text { or } x_{2}=-\sqrt[3]{c}\right) \\
x_{1,3}=\frac{b-a^{2} \pm \sqrt{a^{4}-2 a^{2} b-3 b^{2}}}{2 a}, a \neq 0
\end{gathered}
$$

Now we can directly check the condition:

$$
x_{2}^{2}=x_{1} x_{3}
$$

If $a=0$, then from relation (5) it follows: $b=0$. Therefore, the equation takes the form:

$$
x^{3}+c=0
$$

If $c=0$, then the roots are the numbers $x_{1}=0, x_{2}=0, x_{3}=0$. We can consider that they form a geometric progression with the first term equal to 0 and any common ratio $q$. If $c \neq 0$, then the roots of the equation are three values: $\sqrt[3]{-c}$, $\sqrt[3]{-c} \varepsilon_{1}, \sqrt[3]{-c} \varepsilon_{2}$, where $\varepsilon_{1}=-\frac{1}{2}+i \frac{\sqrt{3}}{2}, \varepsilon_{2}=-\frac{1}{2}-i \frac{\sqrt{3}}{2}$. It is easy to verify that the numbers $\sqrt[3]{-c} \varepsilon_{1}, \sqrt[3]{-c}, \sqrt[3]{-c} \varepsilon_{2}$ form a geometric progression.

Now let $a \neq 0$. Then, if $c=0$, from relation (5) it follows: $b=0$. In this case, the equation has the form:

$$
x^{3}+a x^{2}=0
$$

Its roots are the numbers $-a, 0,0$. We can consider that they form a geometric progression with the first term $a$ and common ratio $q=0$.

Thus, for the roots of the original equation to form a geometric progression, it is necessary and sufficient that the relation $a^{3} c=b^{3}$ be satisfied.