$$\begin{aligned} & x^2-\sqrt{2} x-\sqrt{3}=0 \\ & P_n=\alpha^n-\beta^n \end{aligned}$$

$$\alpha$$ and $$\beta$$ are the roots of the equation

Using Newton's theorem

$$\begin{aligned} & P_{n+2}-\sqrt{2} P_{n+1}-\sqrt{3} P_n=0 \\ & \text { Put } n=10 \\ & P_{12}-\sqrt{2} P_{11}-\sqrt{3} P_{10}=0 \end{aligned}$$

$$P_{12}=\sqrt{2} P_{11}+\sqrt{3} P_{10}$$

Put $$n=9$$

$$\begin{aligned} & P_{11}-\sqrt{2} P_{10}-\sqrt{3} P_9=0 \\ & P_{11}=\sqrt{2} P_{10}+\sqrt{3} P_9 \\ & (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10) P_{11}-11 P_{12} \end{aligned}$$

Put the value of $$P_{12}$$ & $$P_{12}$$ in above equation.

$$\begin{aligned} = & (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10)\left(\sqrt{2} P_{10}+\sqrt{3} P_9\right)-11\left(\sqrt{2} P_{11}+\sqrt{3} P_{10}\right) \\ = & 11 \sqrt{3} P_{10}-10 \sqrt{2} P_{10}+22 P_{10}+10 \sqrt{2} P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-11 \sqrt{2} P_{11}-11 \sqrt{3} P_{10} \\ = & 22 P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-11 \sqrt{2}\left(\sqrt{2} P_{10}+\sqrt{3} P_9\right) \\ = & 22 P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-22 P_{10}-11 \sqrt{6} P_9 \\ = & 10 \sqrt{3} P_9 \end{aligned}$$