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

(N/A) Given: $x = 5 + 2\sqrt{6}$.
First,find $\frac{1}{x}$:
$\frac{1}{x} = \frac{1}{5 + 2\sqrt{6}} = \frac{1}{5 + 2\sqrt{6}} \times \frac{5 - 2\sqrt{6}}{5 - 2\sqrt{6}} = \frac{5 - 2\sqrt{6}}{25 - 24} = 5 - 2\sqrt{6}$.
Now,calculate $x + \frac{1}{x}$:
$x + \frac{1}{x} = (5 + 2\sqrt{6}) + (5 - 2\sqrt{6}) = 10$.
To find $x^{2} + \frac{1}{x^{2}}$:
$x^{2} + \frac{1}{x^{2}} = (x + \frac{1}{x})^{2} - 2 = (10)^{2} - 2 = 100 - 2 = 98$.
To find $x^{3} + \frac{1}{x^{3}}$:
$x^{3} + \frac{1}{x^{3}} = (x + \frac{1}{x})^{3} - 3(x + \frac{1}{x}) = (10)^{3} - 3(10) = 1000 - 30 = 970$.