$\mathbb{Q}(\sqrt[3]{ 2 }) / \mathbb{Q}$ is not Galois because it is not normal as seen in (1) of Example 2209.
In (2) of Exmaple 2209, we know that $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}$ is normal. It is also separable because $\operatorname{ch}(\mathbb{Q}) = 0$ and by Proposition 1704. Thus $\mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q}$ is a Galois extension. Moreover, by (4) of Example 2106, we have$$\mathrm{Gal}\left( \mathbb{Q}(\sqrt[3]{ 2 }, \sqrt{ -3 }) / \mathbb{Q} \right) \cong S_{3}.$$