Correct Answer: 0

Solution : Given: $(x+\frac{1}{x})^{2}=3$,
Taking square root on both sides, we get
$(x+\frac{1}{x})=\sqrt{3}$
Now, Cubing both sides we get
$(x+\frac{1}{x})^3=(\sqrt{3})^3$
⇒ $x^3+\frac{1}{x^3}+3×x×\frac{1}{x}(x+\frac{1}{x})=(3\sqrt{3})$
⇒ $x^3+\frac{1}{x^3}=3\sqrt{3}–3(x+\frac{1}{x})$
$\because x+\frac{1}{x}=\sqrt{3}$
Thus, $x^3+\frac{1}{x^3}=3\sqrt{3}–3\sqrt{3} = 0$
Hence, the correct answer is 0.