Why is $\mathbb{Q}(\sqrt{2})\subseteq\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ In office hours yesterday my instructor said $\mathbb{Q}(\sqrt{2})\subseteq\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$.
I know $\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ 
because $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ is a field so it is closed under addition which means since $\sqrt{2}$ and $\sqrt[3]{5}$ are members then their sum $\sqrt{2}+\sqrt[3]{5}$ is a member.
I am not sure how to show $\mathbb{Q}(\sqrt{2})\subseteq\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$. A friend suggested I could find a linear combination of $\sqrt{2}+\sqrt[3]{5}$ and its inverse that yields $\sqrt{2}$. However I have to admit that I am not sure how to rationalize $\frac{1}{\sqrt{2}+\sqrt[3]{5}}$ in order to even find the inverse of $\sqrt{2}+\sqrt[3]{5}$.
I have found problems similar to this (but simpler) where both members of the sum in question are square roots (instead of one square root and one cube root).
Suggestions?
 A: $K = \mathbb{Q}(\sqrt{2}, \sqrt[3]{5}) $ is seen to have degree 6 over $ \mathbb{Q} $  by the observation that it has subfields of degree 2 and 3. Take the normal closure $ L $ of $ K $ and consider its Galois group. $ K $ has $ 6 $ distinct $\mathbb{Q}$-embeddings into its normal closure, and these embeddings extend to $\mathbb{Q}$-automorphisms of $ L $; which means that $ \sqrt{2} + \sqrt[3]{5} $ has 6 $ \mathbb{Q} $-conjugates. By the Tower Law, we then have 
$$ 6 = [\mathbb{Q}(\sqrt{2}, \sqrt[3]{5}):\mathbb{Q}] =   [\mathbb{Q}(\sqrt{2}, \sqrt[3]{5}):\mathbb{Q}(\sqrt{2} + \sqrt[3]{5})][\mathbb{Q}(\sqrt{2} + \sqrt[3]{5}):\mathbb{Q}] = 6[\mathbb{Q}(\sqrt{2}, \sqrt[3]{5}):\mathbb{Q}(\sqrt{2} + \sqrt[3]{5})]$$
so that $ \mathbb{Q}(\sqrt{2}, \sqrt[3]{5}) = \mathbb{Q}(\sqrt{2} + \sqrt[3]{5})$. The result then follows easily.
A: You know that $$\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$$ and I suppose you know also that the degree of
$$\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$$ is equal to $2\cdot 3=6$. Well, the minimal polynomial of $x=\sqrt{2}+\sqrt[3]{5}$ being of degree $6$ you can ensure that 
  the degree of $x=\sqrt{2}+\sqrt[3]{5}$ is $6$ and consequently $$\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$$ This give you an easy way to verify that 
$$\mathbb{Q}(\sqrt{2})\subseteq\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$$ because obviously $\sqrt 2\in\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})=\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$ 
