Dedekind's Product Theorem (which a friend of mine calls the "Royal Dutch Airlines Theorem") states that if $K\subseteq L\subseteq M$ are fields, then $[M:K] = [M:L][L:K].$ The proof of the theorem is constructive: if $\{m_i\}_{i\in I}$ is a basis for $M$ as an $L$-vector space, and $\{\ell_j\}_{j\in J}$ is a basis for $L$ as a $K$-vector space, then one shows that $\{m_i\ell_j\}_{(i,j)\in I\times J}$ is a basis for $M$ as a $K$-vector space.
The Hint in the book is suggesting that you do this. Since $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4}) = \mathbb{Q}(\sqrt{2},\sqrt[3]{4})$, you can find a basis for $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$ by considering the tower of extensions $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt[3]{4}) \subseteq \mathbb{Q}(\sqrt[3]{4})(\sqrt{2}).$ It is easy to find a minimal polynomial for $\sqrt[3]{4}$ over $\mathbb{Q}$, from which you easily get a basis for the first extension; and it is easy to find a minimal polynomial for $\sqrt{2}$ over $\mathbb{Q}(\sqrt[3]{4})$, from which you get a basis for the second extension. Now you can combine them, using the argument of the Dedekind Product Theorem, to get a basis for $\mathbb{Q}(\sqrt{2},\sqrt[3]{4}) = \mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$ over $\mathbb{Q}$.