Solution. Let $a, a q, a q^{2}$ be the roots of the original equation. Then, by Vieta's theorem, $\left\{\begin{array}{l}a^{3} q^{3}=-\frac{1}{64}, \\ a+a q+a q^{2}=\frac{3}{8}\end{array} \Leftrightarrow\left\{\begin{array}{l}a q=-\frac{1}{4}, \\ 2 q^{2}+5 q+2=0\end{array} \Leftrightarrow\left[\begin{array}{l}q=-2, a=\frac{1}{8}, \\ q=-\frac{1}{2}, a=\frac{1}{2} .\end{array}\right.\right.\right.$

Answer: The roots of the equation are: $\frac{1}{8} ;-\frac{1}{4} ; \frac{1}{2}$.
6.363 Solve the equation $64 x^{3}-24 x^{2}-6 x+1=0$, given that its roots form a geometric progression.