Solution: Since $x^3 - \sqrt{3} $ is a polynomial of degree $3$, so it is reducible over $\mathbb{Q}\left( \sqrt{3} \right) $ if and only if the polynomial has a root in $\mathbb{Q} \left( \sqrt{3} \right) $. We note that $x^3 - 3$ is irreducible over $\mathbb{Q} $. Therefore, if $\alpha $ is a root of $x^3 - 3$, then the degree of $\mathbb{Q} (\alpha )$ over $\mathbb{Q} $ must be $3$.

Suppose that the polynomial $x^3 - 3$ is reducible over $\mathbb{Q} (\sqrt{3})$. Then there must be a root of $x^3 - 3$ in $\mathbb{Q} \left( \sqrt{3} \right) $, say $\alpha $. So we have \begin{align*} 2 = \left[ \mathbb{Q} \left( \sqrt{3} \right) , \mathbb{Q} \right] = \left[ \mathbb{Q} \left( \sqrt{3} \right) , \mathbb{Q} (\alpha ) \right] \left[ \mathbb{Q} (\alpha ), \mathbb{Q} \right] \geq 3, \end{align*} which is a contradiction. Hence the polynomial is irreducible over $\mathbb{Q} \left( \sqrt{3} \right) $.