21.5. (Austria, 83). Find all values of $a$ for which the roots $x_{1}, x_{2}, x_{3}$ of the polynomial $x^{3}-6 x^{2}+a x+a$ satisfy the equation

$$
\left(x_{1}-3\right)^{3}+\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0
$$
21.5. Let's make the substitution $y=x-3$, then the numbers $y_{1}=x_{1}-3, y_{2}=$ $=x_{2}-3$ and $y_{3}=x_{3}-3$ are the roots of the polynomial

$$
(y+3)^{3}-6(y+3)^{2}+a(y+3)+a=y^{3}+3 y^{2}+(a-9) y+4 a-27
$$

By Vieta's theorem, we have the equalities

$$
\begin{aligned}
& y_{1}+y_{2}+y_{3}=-3 \\
& y_{1} y_{2}+y_{1} y_{3}+y_{2} y_{3}=a-9 \\
& y_{1} y_{2} y_{3}=27-4 a
\end{aligned}
$$

and in addition, the relation $y_{1}^{3}+y_{2}^{3}+y_{3}^{3}=0$ must hold. By direct verification, we confirm the validity of the identity $y_{1}^{3}+y_{2}^{3}+y_{3}^{3}=$

$$
\equiv\left(y_{1}+y_{2}+y_{3}\right)^{3}-3\left(y_{1} y_{2}+y_{1} y_{3}+y_{2} y_{3}\right)\left(y_{1}+y_{2}+y_{3}\right)+3 y_{1} y_{2} y_{3}
$$

from which we obtain the necessary and sufficient condition for $a$ :

$$
0=(-3)^{3}-3(a-9) \cdot(-3)+3(27-4 a)=-27-3 a
$$

i.e. $a=-9$.