Question

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

Answer 1

Given that,

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

On cube both side and we get ${(x+\frac{1}{x})}^3={\left(\sqrt{3}\right)}^3$

$\Rightarrow x^3+\frac{1}{x^3}+3x.\frac{1}{x}(x+\frac{1}{x})=3\sqrt{3}$

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

$\Rightarrow x^3+\frac{1}{x^3}=0$

Hence, this is the answer.