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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 $52$.