Let $\alpha, \beta ; \alpha>\beta$, be the roots of the equation $x^2-\sqrt{2} x-\sqrt{3}=0$. Let $\mathrm{P}_n=\alpha^n-\beta^n, n \in \mathrm{N}$. Then $$(11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}$$ is equal to

<p>$$\begin{aligned} & x^2-\sqrt{2} x-\sqrt{3}=0 \\ & P_n=\alpha^n-\beta^n \end{aligned}$$</p> <p>$\alpha$ and $\beta$ are the roots of the equation</p> <p>Using Newton's theorem</p> <p>$$\begin{aligned} & P_{n+2}-\sqrt{2} P_{n+1}-\sqrt{3} P_n=0 \\ & \text { Put } n=10 \\ & P_{12}-\sqrt{2} P_{11}-\sqrt{3} P_{10}=0 \end{aligned}$$</p> <p>$P_{12}=\sqrt{2} P_{11}+\sqrt{3} P_{10}$</p> <p>Put $n=9$</p> <p>$$\begin{aligned} & P_{11}-\sqrt{2} P_{10}-\sqrt{3} P_9=0 \\ & P_{11}=\sqrt{2} P_{10}+\sqrt{3} P_9 \\ & (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10) P_{11}-11 P_{12} \end{aligned}$$</p> <p>Put the value of $P_{12}$ & $P_{12}$ in above equation.</p> <p>$$\begin{aligned} = & (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10)\left(\sqrt{2} P_{10}+\sqrt{3} P_9\right)-11\left(\sqrt{2} P_{11}+\sqrt{3} P_{10}\right) \\ = & 11 \sqrt{3} P_{10}-10 \sqrt{2} P_{10}+22 P_{10}+10 \sqrt{2} P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-11 \sqrt{2} P_{11}-11 \sqrt{3} P_{10} \\ = & 22 P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-11 \sqrt{2}\left(\sqrt{2} P_{10}+\sqrt{3} P_9\right) \\ = & 22 P_{10}+11 \sqrt{6} P_9+10 \sqrt{3} P_9-22 P_{10}-11 \sqrt{6} P_9 \\ = & 10 \sqrt{3} P_9 \end{aligned}$$</p>