smc

We know $r+\frac{1}{r}=\sqrt{3}$. Cubing this gives $r^3+3r+\frac{3}{r}+\frac{1}{r^3}=3\sqrt{3}$. But $3r+\frac{3}{r}=3\left(r+\frac{1}{r}\right)=3\sqrt{3}$, so subtracting this from the first equation gives $r^3+\frac{1}{r^3}=\boxed{0}$. (Actually, $r+\frac{1}{r}$ could have been equal to $-\sqrt{3}$ instead of $\sqrt{3}$, but this would have led to the same answer. Also, this answer implies that $r^6=-1$, which means that $r$ is a complex number.)