As $9\mid P-1$, $9\mid P^2-1$ so we may take the an element $t$ of $\mathbb{F}_{P^2}$ with order 9. Then, note that $t+\frac1t\in\mathbb{F}_P$ and $\left(t+\frac1t\right)^3-3\left(t+\frac1t\right)+1=t^3+\frac1{t^3}+1=\frac{t^6+t^3+1}{t^3}=0$, so we may take $n\equiv t+\frac1t\pmod{P}$.