Problem 7. For what values of the parameter $a$ does the equation $x^{3}-15 x^{2}+a x-64=0$ have three distinct real roots that form a geometric progression?
Answer: only 60.

Solution. Let the parameter $a$ be suitable. Then the polynomial $x^{3}-15 x^{2}+a x-64=0$ has three distinct roots $x_{1}, x_{2}, x_{3}$. We use Vieta's theorem for a polynomial of the third degree:

$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=15 \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=a \\
x_{1} x_{2} x_{3}=64
\end{array}\right.
$$

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 $4\left(\frac{1}{q}+1+q\right)=15(*)$, after transformations $4 q^{2}-11 q+4=0$. The discriminant of this expression is $D=11^{2}-4 \cdot 4 \cdot 4>0$, so such $q$, and with it $x_{1}, x_{2}, x_{3}$, will exist. Then

$a=x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=x_{1} x_{2} x_{3}\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right)=64 \cdot\left(\frac{q}{4}+\frac{1}{4}+\frac{1}{4 q}\right)=16\left(q+1+\frac{1}{q}\right)=4 \cdot 15=60$. (In the penultimate transition, we used the equality $(*)$ ).

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

$$
q_{1,2}=\frac{11 \pm \sqrt{57}}{8}
$$

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

$$
x_{1}=\frac{4 \cdot 8}{11+\sqrt{57}}=\frac{11-\sqrt{57}}{2}, x_{2}=4, x_{3}=\frac{11+\sqrt{57}}{2} .
$$

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