<s>[INST] Prove that $\Bbb Q(\sqrt{2},3^{1/3})=\Bbb Q(\sqrt{2}+3^{1/3})$ 
Prove that $\Bbb Q(\sqrt{2},3^{1/3})=\Bbb Q(\sqrt{2}+3^{1/3})$

My attempt:
Firstly, since, $\Bbb Q(\sqrt{2}+3^{1/3}) \subseteq \Bbb Q(\sqrt{2},3^{1/3})$ , I computed $(\sqrt{2}+3^{1/3})^{-1} = 6-4\sqrt{2}+4(3^{1/3})+3(3^{2/3})-3(\sqrt{2}3^{1/3})-2(\sqrt{2}3^{2/3})$ and also tried some naive manipulation by considering powers of $(\sqrt{2}+3^{1/3})$ and tried to eliminate terms ( imitating the proof for $\Bbb Q(\sqrt{2},\sqrt{3})=\Bbb Q(\sqrt{2}+\sqrt{3})$ ) but it gets very messy.
Secondly, I am trying this problem by Galois theory but since, $\Bbb Q(\sqrt{2},3^{1/3})|\Bbb Q$ is not Galois,I don't get how to proceed.
Thanks in advance for help!
 [/INST] Let $\alpha = \sqrt2+ \sqrt[3]{3}$.
First isolate $\sqrt[3]{3}$ and raise everything to the third power:
$$\sqrt[3]{3} = \alpha - \sqrt{2} \implies 3 = (\alpha - \sqrt{2})^3 = \alpha^3-3\alpha^2\sqrt{2}+6\alpha - 2\sqrt{2}$$
Then isolate $\sqrt{2}$
$$\sqrt{2}(3\alpha^2+2) = \alpha^3+6\alpha-3$$
so $$\sqrt{2} = \frac{\alpha^3+6\alpha-3}{3\alpha^2+2} \in \mathbb{Q}(\alpha)$$
Then also $\sqrt[3]{3} = \alpha - \sqrt{2} \in \mathbb{Q}(\alpha)$ so $\mathbb{Q}(\alpha) = \mathbb{Q}(\sqrt2, \sqrt[3]{3})$.
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