<s>[INST] Is $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})=\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$? I understand that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2},\sqrt[3]{5})$
But I am struggling to algebraically show that $\sqrt{2}$,$\sqrt[3]{5}\in\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$ to conclude that $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})\subseteq\mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$
 [/INST] I'm afraid I can't think of any methods other than brute force.  If we set $\alpha = \sqrt{2} + \sqrt[3]{5}$, then all powers of $\alpha$ lie in the span over $\Bbb Q$ of the set $\{1, \sqrt{2}, \sqrt[3]{5}, \sqrt[3]{25}, \sqrt{2}\sqrt[3]{5}, \sqrt{2}\sqrt[3]{25}\}.$  What we essentially need to show is that the set spanned by the powers of $\alpha$ then contains $6$ linearly independent elements of this span.
I start by calculating $\alpha^2 = 2 + \sqrt[3]{25} + 2\sqrt{2}\sqrt[3]{5}$.  You could then compute $\alpha^3, \alpha^4, \alpha^5$ similarly, and check by row reduction that the resulting set is linearly independent.  
If you're doing computations by hand, it might be somewhat easier, to compute $\alpha(\alpha^2-2) = (\sqrt{2} + \sqrt[3]{5})(\sqrt[3]{25} + 2\sqrt{2}\sqrt[3]{5}) = 5 + 4\sqrt[3]{5} + 3\sqrt{2}\sqrt[3]{25}$, and then similarly compute $\alpha(\alpha(\alpha^2-2)-5)$, and so forth.
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