In Dummit and Foote, Abstract Algebra, Sec 13.4 Example 3, The splitting field of $x^{3}-2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[3]{2}, \sqrt{-3})$. Then he says, quote : Since $\sqrt{-3}$ satisfies the equation $x^{2}+3 =0$, the degree of this extension over $\mathbb{Q}$ is at most 2, hence must be 2 since we observe that $\mathbb{Q}(\sqrt[3]{2})$ is not the splitting field. 

I do not understand why he considers this. Could somebody be more specific and explain me why this is so?

Thanks