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

Given:

\( x=2+\sqrt{3} \)

To do: 

We have to find the value of \( x^{3}+\frac{1}{x^{3}} \).

Solution:

We know that,

Rationalising factor of a fraction with denominator ${\sqrt{a}}$ is ${\sqrt{a}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}-\sqrt{b}}$ is ${\sqrt{a}+\sqrt{b}}$.

Rationalising factor of a fraction with denominator ${\sqrt{a}+\sqrt{b}}$ is ${\sqrt{a}-\sqrt{b}}$.

Therefore,

$x=2+\sqrt{3}$

$\Rightarrow \frac{1}{x}=\frac{1}{2+\sqrt{3}}$

$=\frac{1(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}$

$=\frac{2-\sqrt{3}}{(2)^{2}-(\sqrt{3})^{2}}$

$=\frac{2-\sqrt{3}}{4-3}$

$=2-\sqrt{3}$

Therefore,

$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4$

Cubing both sides, we get,

$(x+\frac{1}{x})^{3}=(4)^{3}$

$x^{3}+\frac{1}{x^{3}}+3(x+\frac{1}{x})=64$

$\Rightarrow x^{3}+\frac{1}{x^{3}}+3\times 4=64$

$\Rightarrow x^{3}+\frac{1}{x^{3}}=64-12$

$\Rightarrow x^{3}+\frac{1}{x^{3}}=52$

The value of \( x^{3}+\frac{1}{x^{3}} \) is $52$.