5.75 Find all values of $a$ such that the polynomial
$$x^{3}-6 x^{2}+a x+a$$

has roots $x_{1}, x_{2}, x_{3}$ satisfying
$$\left(x_{1}-3\right)^{3}+\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0$$
[Solution] Make the variable substitution $y=x-3$, then $y_{1}=x_{1}-3, y_{2}=x_{2}-3$ and $y_{3}=x_{3}-3$ are the roots of the polynomial
$$\begin{aligned}
& (y+3)^{3}-6(y+3)^{2}+a(y+3)+a \\
= & y^{3}+3 y^{2}+(a-9) y+4 a-27
\end{aligned}$$

By Vieta's formulas, we have
$$\left\{\begin{array}{l}
y_{1}+y_{2}+y_{3}=-3 \\
y_{1} y_{2}+y_{1} y_{3}+y_{2} y_{3}=a-9 \\
y_{1} y_{2} y_{3}=27-4 a
\end{array}\right.$$

At this point, by the assumption, we also have
$$y_{1}^{3}+y_{2}^{3}+y_{3}^{3}=0$$

It is not difficult to verify that the following identity holds:
$$\begin{aligned}
y_{1}^{3}+y_{2}^{3}+y_{3}^{3} \equiv & \left(y_{1}+y_{2}+y_{3}\right)^{3}-3\left(y_{1} y_{2}+y_{2} y_{3}+y_{1} y_{3}\right) \\
& \left(y_{1}+y_{2}+y_{3}\right)+3 y_{1} y_{2} y_{3}
\end{aligned}$$

From this, we get that the necessary and sufficient condition for $a$ to meet the requirements of the problem is
$$\begin{aligned}
0 & =(-3)^{3}-3(a-9)(-3)+3(27-4 a) \\
& =-27-3 a
\end{aligned}$$

Therefore, $\quad a=-9$.