9. Find all values of $a$ such that the roots $x_{1}, x_{2}, x_{3}$ of the polynomial $x^{3}-6 x^{2}+a x+a$ satisfy $\left(x_{1}-3\right)^{3}+\left(x_{2}-3\right)^{3}+\left(x_{3}-3\right)^{3}=0$.
9. Let the variable substitution be $y=x-3$. By Vieta's formulas, and given that $y_{1}^{3}+y_{2}^{3}+y_{3}^{3}=0$, we solve to get $a=-9$.