If $x=5+2 \sqrt{6},$ then find the value of $x^{2}+\frac{1}{x^{2}}$ and $x^{3}+\frac{1}{x^{3}}$

Hint $:\left(x+\frac{1}{x}\right)^{2}=x^{2}+2+\frac{1}{x^{2}}$ and $\left(x+\frac{1}{x}\right)^{3}=x^{3}+\frac{1}{x^{3}}+3(x)\left(\frac{1}{x}\right)\left(x+\frac{1}{x}\right)$

$x=5+2 \sqrt{6} \quad$ (Given)

$\therefore \frac{1}{x}=\frac{1}{5+2 \sqrt{6}}$

$=\frac{1}{5+2 \sqrt{6}} \times \frac{5-2 \sqrt{6}}{5-2 \sqrt{6}}$

$=\frac{1[5-2 \sqrt{6}]}{(5)^{2}-(2 \sqrt{6})^{2}}$

$=\frac{5-2 \sqrt{6}}{25-24}$

$=5-2 \sqrt{6}$

$\therefore x+\frac{1}{x}=(5+2 \sqrt{6})+(5-2 \sqrt{6})=10$

Now, $x^{2}+\frac{1}{x^{2}}=\left(x+\frac{1}{x}\right)^{2}-2$

$=(10)^{2}-2=100-2=98$

$x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)^{3}-3(x)\left(\frac{1}{x}\right)\left(x+\frac{1}{x}\right)$

$=(10)^{3}-3(10)$

$=1000-30=970$