473. Solve the cubic equation:

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

if its roots form:

1) an arithmetic progression;
2) a geometric progression;
3) a harmonic series.
473. It is known that between the roots $x_{1}, x_{2}, x_{3}$ and the coefficients $a, b, c$ of the cubic equation $x^{3}+a x^{2}+b x+c=0$, there exists a dependency:

$$
\begin{gathered}
a=-\left(x_{1}+x_{2}+x_{3}\right) \\
b=x_{1} x_{2}+x_{2} x_{3}+x_{1} x_{3} \\
c=-x_{1} x_{2} x_{3}
\end{gathered}
$$

1) If the roots form an arithmetic progression, we have an additional condition:

$$
x_{1}-x_{2}=x_{2}-x_{3},
$$

then from (1) and (4)

$$
x_{2}=-\frac{a}{3} ;
$$

from (1) and (3)

$$
\begin{gathered}
x_{1}+x_{3}=-\frac{2}{3} a \\
x_{1} x_{3}=\frac{3 c}{a}
\end{gathered}
$$

from which

$$
\begin{aligned}
& x_{1}=-\frac{a}{3}+\frac{1}{3} \sqrt{\frac{a^{3}-27 c}{a}} \\
& x_{3}=-\frac{a}{3}-\frac{1}{3} \sqrt{\frac{a^{3}-27 c}{a}}
\end{aligned}
$$

If the found roots are substituted into (2), we obtain the condition that the coefficients must satisfy for the cubic equation to have roots forming an arithmetic progression. This condition will be:

$$
\frac{2 a^{3}+27 c}{9 a}=b
$$

Conversely, if there is such a relationship between the coefficients, the roots of the equation will be terms of an arithmetic progression.

2) If the roots of the equation form a geometric progression, an additional condition is added:

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

From (5) and (3) it follows:

$$
x_{2}=\sqrt[3]{-c} ; x_{1} x_{3}=\sqrt[3]{c^{2}} ;
$$

from (1):

$$
x_{1}+x_{3}=-a+\sqrt[3]{c}
$$

from (6) and (7) we get:

$$
\begin{aligned}
& x_{1}=-\frac{a-\sqrt[3]{c}}{2}-\sqrt{\left(\frac{a-\sqrt[3]{c}}{2}\right)^{2}-\sqrt[3]{c^{2}}} \\
& x_{3}=-\frac{a-\sqrt[3]{c}}{2}+\sqrt{\left(\frac{a-\sqrt[3]{c}}{2}\right)^{2}-\sqrt[3]{c^{2}}}
\end{aligned}
$$

Substituting $x_{1}, x_{2}, x_{3}$ into (2), we find the condition:

$$
b=a \sqrt[3]{c}
$$

under which the roots of the cubic equation form a geometric progression. Conversely, if this condition is met, the roots will necessarily be terms of a geometric progression.

3) If the roots form a harmonic series, an additional condition is added:

$$
\frac{x_{1}-x_{2}}{x_{2}-x_{3}}=\frac{x_{1}}{x_{3}}
$$

or

$$
2 x_{1} x_{3}=x_{2}\left(x_{1}+x_{3}\right) \text {. }
$$

From (2) and (8) we get:

$$
x_{1} x_{3}=\frac{b}{3}
$$

and from (3):

$$
x_{2}=-\frac{3 c}{b}
$$

Substituting into (8), we find:

$$
\frac{2}{3} b=-\frac{3 c}{b}\left(x_{1}+x_{3}\right)
$$

From (9) and (10):

$$
\begin{aligned}
& x_{1}=-\frac{b^{2}}{9 c}+\sqrt{\left(\frac{b^{2}}{9 c}\right)^{2}-\frac{b}{3}} \\
& x_{3}=-\frac{b^{2}}{9 c}-\sqrt{\left(\frac{b^{2}}{9 c}\right)^{2}-\frac{b}{3}}
\end{aligned}
$$

Substituting $x_{1}, x_{2}, x_{3}$ into (1), we obtain the condition:

$$
\frac{2 b^{3}+27 c}{9 b c}=a
$$

under which the roots form a harmonic series.