1. We are given that $x + \frac{1}{x} = \sqrt{3}$. Our goal is to find the value of $x^3 + \frac{1}{x^3}$. 2. Recall the identity for cubes: $$x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right)$$ 3. Substitute $x + \frac{1}{x} = \sqrt{3}$ into the identity: $$x^3 + \frac{1}{x^3} = (\sqrt{3})^3 - 3(\sqrt{3})$$ 4. Calculate each term: $$(\sqrt{3})^3 = (3^{1/2})^3 = 3^{3/2} = 3 \sqrt{3}$$ 5. Substitute back: $$x^3 + \frac{1}{x^3} = 3\sqrt{3} - 3\sqrt{3} = 0$$ 6. Thus, the value of $x^3 + \frac{1}{x^3}$ is $0$.