4. Find all values of $a$ such that for each of its values, the equation
$$
x^{3}-a x^{2}-\left(a^{3}-6 a^{2}+5 a+8\right) x-(a-3)^{3}=0
$$

has three distinct roots, which form a geometric progression, and find these roots.
4. Let $x_{1}, x_{2}, x_{3}$ be the roots of the equation. By Vieta's formulas, we have
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}=a, \\
x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=-a^{3}+6 a^{2}-5 a-8, \\
x_{1} x_{2} x_{3}=(a-3)^{3} .
\end{array}
$$

Given that $x_{2}^{2}=x_{1} x_{3}$.
Substituting into (3) gives $x_{2}=a-3$.
Substituting into (1) and (3) gives
$$
x_{1}+x_{3}=3, x_{1} x_{3}=(a-3)^{2} \text {. }
$$

Substituting into (2) and simplifying gives
$$
a^{3}-5 a^{2}+2 a+8=0 \Leftrightarrow a=-1,2,4 \text {. }
$$

When $a=-1$, $x_{2}=-4$, and $x_{1}$ and $x_{3}$ have no real solutions;
When $a=2$, $x_{2}=-1$, $x_{1}=\frac{3-\sqrt{5}}{2}$, $x_{3}=\frac{3+\sqrt{5}}{2}$;
When $a=4$, $x_{2}=1$, $x_{1}=\frac{3+\sqrt{5}}{2}$, $x_{3}=\frac{3-\sqrt{5}}{2}$.
Thus, when $a=2$, the roots are $\frac{3-\sqrt{5}}{2},-1, \frac{3+\sqrt{5}}{2}$;
When $a=4$, the roots are $\frac{3-\sqrt{5}}{2}, 1, \frac{3+\sqrt{5}}{2}$.