If x = 2+√3, find the value of

Solution:

Given, $x=2+\sqrt{3}$

To find the value of $x^{3}+\frac{1}{x^{3}}$

We have, $x=2+\sqrt{3}$,

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

Rationalizing the denominator by multiplying both numerator and denominator with the rationalizing factor

$2-\sqrt{3}$ for $\frac{1}{2+\sqrt{3}}$

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

Since, $(a+b)(a-b)=\left(a^{2}-b^{2}\right)$

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

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

$x+\frac{1}{x}=4$

We know that, $\left(a^{3}+b^{3}\right)=(a+b)\left(a^{2}-a b+b^{2}\right)$

$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(x^{2}-x \cdot \frac{1}{x}+\frac{1}{x^{2}}\right)$

$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1^{2}}{x}-1\right)$

$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}+2-2-1\right)$

$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(x^{2}+\frac{1}{x^{2}}+2\left(x \cdot \frac{1}{x}\right)-2-1\right)$

$\left(x^{3}+\frac{1}{x^{3}}\right)=\left(x+\frac{1}{x}\right)\left(\left(x+\frac{1}{x}\right)^{2}-3\right)$

Putting the value of $x+1 x$ in the above equation, we get,

$\left(x^{3}+\frac{1}{x^{3}}\right)=(4)\left(4^{2}-3\right)$

$\left(x^{3}+\frac{1}{x^{3}}\right)=52$