This is exercise A4 in chapter 29 from Pinter's A Book of Abstract Algebra. It is not homework but hints/roadmap would be preferred to a full solution for now.
First some context
The book works out the example of $\mathbb{Q}(\sqrt{1+\sqrt[3]{2}})$. If $a = \sqrt{1+\sqrt[3]{2}}$ then $a^2-1 = \sqrt[3]{2}$, so that $\sqrt[3]{2} \in \mathbb{Q}(a)$, from which it follows that $\mathbb{Q}(a) = \mathbb{Q}(\sqrt[3]{2},a)$. We then adjoin $\sqrt[3]{2}$ and $a$ to $\mathbb{Q}$ in order.
The field $\mathbb{Q}(\sqrt[3]{2})$ is an extension of degree $3$ over $\mathbb{Q}$ with basis $\{1,2^{1/3},2^{2/3}\}$. Since $x^2-1-\sqrt[3]{2}$ is irreducible over $\mathbb{Q}(\sqrt[3]{2})$, the field $\mathbb{Q}(\sqrt[3]{2},a)$ is an extension of degree $2$ over $\mathbb{Q}(\sqrt[3]{2})$ with basis $\{1,a\}$. Thus $\mathbb{Q}(\sqrt[3]{2},a)$ is an extension of degree $6$ over $\mathbb{Q}$ with basis $\{1,2^{1/3},2^{2/3},a,a 2^{1/3},a 2^{2/3}\}$.
The question at hand
I have to find a basis for the field extension $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$.
A hint is given: This is similar to the case for $\mathbb{Q}(\sqrt{1+\sqrt[3]{2}})$. Adjoin first $\sqrt[3]{4}$, then $a$.
I'm not sure if the $a$ in the hint refers to the same $a$ as in the example or if it refers to $\sqrt{2}+\sqrt[3]{4}$.
Naively I computed
$ \left(\sqrt{2}+\sqrt[3]{4}\right)^2 = 2 + 2 \sqrt{2} \sqrt[3]{4} + \left(\sqrt[3]{4}\right)^2 $
and
$ \left(\sqrt{2}+\sqrt[3]{4}\right)^3 = 4 + 2\sqrt{2} + 6 \sqrt[3]{4} + 3\sqrt{2}\left(\sqrt[3]{4}\right)^2, $
but I'm not seeing how to combine these to show that $\sqrt[3]{4} \in \mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$, which would allow me to follow the steps of the example.