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

(N/A) Given $x = 2 + \sqrt{3}.$
First,find $\frac{1}{x} = \frac{1}{2 + \sqrt{3}}.$
Rationalizing the denominator: $\frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}.$
Now,$x + \frac{1}{x} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4.$
To find $x^{2} + \frac{1}{x^{2}},$ use the identity $(x + \frac{1}{x})^{2} = x^{2} + \frac{1}{x^{2}} + 2.$
$4^{2} = x^{2} + \frac{1}{x^{2}} + 2 \implies 16 = x^{2} + \frac{1}{x^{2}} + 2 \implies x^{2} + \frac{1}{x^{2}} = 14.$
To find $x^{3} + \frac{1}{x^{3}},$ use the identity $(x + \frac{1}{x})^{3} = x^{3} + \frac{1}{x^{3}} + 3(x + \frac{1}{x}).$
$4^{3} = x^{3} + \frac{1}{x^{3}} + 3(4) \implies 64 = x^{3} + \frac{1}{x^{3}} + 12 \implies x^{3} + \frac{1}{x^{3}} = 52.$
Thus,the values are $14$ and $52.$