5. Find the conditions on the coefficients of the polynomial $x^{3}+a x^{2}+b x+c$, under which this polynomial has three distinct real roots that are consecutive terms of some geometric progression.
6.I.5. Let $x_{1}, x_{2}, x_{3}$ be the roots of the polynomial, which are consecutive terms of a geometric progression. Since $x_{2}^{3}=x_{1} x_{2} x_{3} = -c$, then $x_{2}=-\sqrt[3]{c}$. Substituting the found value into the given polynomial, we get $-c + a \sqrt[3]{c^{2}} - b \sqrt[3]{c} + c = 0$, from which $a \sqrt[3]{c^{2}} = b \sqrt[3]{c}$. Since $x_{2} \neq 0$, it follows that $a^{3} c = b^{3}$.

Now suppose that $a^{3} c = b^{3}$, and find the additional condition under which the polynomial will have three real roots forming a geometric progression. Note that the numbers $a$ and $b$ are zero simultaneously, and the equation $x^{3} + c = 0$ has only one real root. Therefore, $a \neq 0$ and $b \neq 0$. We have,

\[
\begin{aligned}
x^{3} + a x^{2} + b x + c = x^{3} + \frac{b^{3}}{a^{3}} + & a x \left(x + \frac{b}{a}\right) = \\
& = \left(x + \frac{b}{a}\right) \left(x^{2} + \left(a - \frac{b}{a}\right) x + \frac{b^{2}}{a^{2}}\right)
\end{aligned}
\]

Let $x_{2} = -\frac{b}{a}$. We will find the conditions under which the quadratic trinomial in the parentheses has real roots, for which we will find its discriminant. We have,

\[
\begin{aligned}
D = \left(a - \frac{b}{a}\right)^{2} - \frac{4 b^{2}}{a^{2}} = a^{2} - & 2 b - \frac{3 b^{2}}{a^{2}} = \\
= & \frac{a^{4} - 2 a^{2} b - 3 b^{2}}{a^{2}} = -\frac{\left(3 b - a^{2}\right)\left(b + a^{2}\right)}{a^{2}}
\end{aligned}
\]

Therefore, the polynomial has two distinct roots if and only if the inequalities $-a^{2} < b < \frac{a^{2}}{3}$ hold. It remains to note that if $x_{1}$ and $x_{3}$ are the roots of this trinomial, then $x_{1} x_{3} = \frac{b^{2}}{a^{2}} = x_{2}^{2}$, so the numbers $x_{1}, x_{2}$, and $x_{3}$ form a geometric progression.

Answer: $a^{3} c = b^{3}$, with $-a^{2} < b < \frac{a^{2}}{3}$.