If \(f\left(x+\frac{1}{x}\right)=x^3+\frac{1}{x^3}\) then \(f(\sqrt{3})\) is equal to

\(0\) \(1\) \(\sqrt{3}\) \(3\sqrt{3}\)

\(0\)

$f\left(x+\frac{1}{x}\right)=\frac{1}{x^3}+x^3$ $=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)$ so $f(y)=y^3-3y$ at $y=\sqrt{3}$ $f(\sqrt{3})={\sqrt{3}}^2-3\sqrt{3}=0$