9.222 For what values of $p$ is the inequality $-9<\frac{3 x^{2}+p x-6}{x^{2}-x+1}<6$ satisfied for all values of $x$?
Solution. The inequality is equivalent to the following inequality:
$-9\left(x^{2}-x+1\right)<0, \\ 3 x^{2}-(p+6) x+12>0 .\end{array}\right.$

Since the solution to this system should be the set $(-\infty ;+\infty)$, each inequality must be satisfied for all $x$, i.e., the discriminants $D_{1}$ and $D_{2}$ are such that:

$\left\{\begin{array}{l}D_{1}<0, \\ D_{2}<0\end{array} \Leftrightarrow\left\{\begin{array}{l}(p-9)^{2}-144<0, \\ (p+6)^{2}-144<0\end{array} \Leftrightarrow\left\{\begin{array}{l}(p-21)(p+3)<0, \\ (p-6)(p+18)<0\end{array} \Leftrightarrow-3<p<6\right.\right.\right.$.

Answer: $p \in(-3 ; 6)$.