If f≤ft(x+1/x)=x^3+1/(x^3) then f(√3) is equal to

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

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

so $f(y)=y^3-3y$

at $y=\sqrt{3}$

$f(\sqrt{3})={\sqrt{3}}^2-3\sqrt{3}=0$