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

(N/A) Given $x = 3 + 2\sqrt{2}$.
First,find $\frac{1}{x}$:
$\frac{1}{x} = \frac{1}{3 + 2\sqrt{2}} = \frac{1}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{9 - 8} = 3 - 2\sqrt{2}$.
Now,calculate $x + \frac{1}{x}$:
$x + \frac{1}{x} = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6$.
To find $x^{2} + \frac{1}{x^{2}}$,use the identity $(x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x^{2}} + 2$:
$x^{2} + \frac{1}{x^{2}} = (x + \frac{1}{x})^{2} - 2 = (6)^{2} - 2 = 36 - 2 = 34$.
To find $x^{3} + \frac{1}{x^{3}}$,use the identity $(x + \frac{1}{x})^{3} = x^{3} + \frac{1}{x^{3}} + 3(x + \frac{1}{x})$:
$x^{3} + \frac{1}{x^{3}} = (x + \frac{1}{x})^{3} - 3(x + \frac{1}{x}) = (6)^{3} - 3(6) = 216 - 18 = 198$.
Thus,the values are $34$ and $198$.