Problem 7. For what values of the parameter $a$ does the equation $x^{3}+16 x^{2}+a x+64=0$ have three distinct real roots that form a geometric progression?
Answer: only 64.

Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}+16 x^{2}+a x+64=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a cubic polynomial:

$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=-16 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2} x_{3}=-64
\end{array}\right.
$$

Since $x_{1}, x_{2}, x_{3}$ form a geometric progression (let's assume in this order), there exist $b$ and $q$ such that $x_{1}=b, x_{2}=b q, x_{3}=b q^{2}$. Then from the equality $x_{1} x_{2} x_{3}=-64$, we have $b^{3} q^{3}=-64$, from which $x_{2}=b q=-4, x_{1}=-4 / q, x_{3}=-4 q$.

Then $-4\left(\frac{1}{q}+1+q\right)=-16(*)$, after transformations $q^{2}-3 q+1=0$. The discriminant of this expression is $D=3^{2}-4 \cdot 1 \cdot 1>0$, so such a $q$, and with it $x_{1}, x_{2}, x_{3}$, will exist. Then

$a=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=x_{1} x_{2} x_{3}\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right)=-64 \cdot\left(-\frac{q}{4}-\frac{1}{4}-\frac{1}{4 q}\right)=16\left(q+1+\frac{1}{q}\right)=4 \cdot 16=64$.

(In the penultimate transition, we used the equality $\left({ }^{*}\right)$ ).

Comment. Naturally, $q, x_{1}, x_{2}, x_{3}$ can be calculated explicitly:

$$
q_{1,2}=\frac{3 \pm \sqrt{5}}{2}
$$

Choose $q$ with the “+” (if you choose with the “-”, then $x_{1}$ and $x_{3}$ will swap places, which will not affect the answer); then

$$
x_{1}=\frac{-4 \cdot 2}{3+\sqrt{5}}=-2(3-\sqrt{5}), x_{2}=-4, x_{3}=-2(3+\sqrt{5})
$$

$a$ could have been calculated by substituting the obtained numbers into the expression $x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}$.