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I am doing some early study in field theory and am stuck on the following problem.

Show that $\mathbb{Q}(\sqrt{2}) \subseteq \mathbb{Q}(\sqrt{2}+\sqrt[3]{2})$ and that $\mathbb{Q}(\sqrt[3]{2}) \subseteq \mathbb{Q}(\sqrt{2}+\sqrt[3]{2})$, and hence deduce that $\mathbb{Q}(\sqrt{2}+\sqrt[3]{2}) = \mathbb{Q}(\sqrt{2},\sqrt[3]{2})$.

My initial thoughts were to use the fact that $\mathbb{Q}(\sqrt{2})$ must be the smallest field containing $\mathbb{Q}$ as a subfield and with $\sqrt{2}$ (likewise a similar process for the other inclusion), but can't seem to make meaningful progress with this approach. More specifically, I don't know how to show that $\sqrt{2} \in \mathbb{Q}(\sqrt{2}+\sqrt[3]{2})$.

Any help would be great!

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    $\begingroup$ Let $\alpha = \sqrt2+\sqrt[3]2$. Take a rational combination of $\alpha, \alpha^2, \alpha^3$ that eliminates the terms $\sqrt[3]2$ and $\sqrt[3]2^2$. Likewise $\sqrt2$ can be eliminated from $\alpha, \alpha^2$. $\endgroup$ – WimC Sep 2 at 18:52
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    $\begingroup$ A more difficult generalization. $\endgroup$ – Jyrki Lahtonen Sep 2 at 18:59
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Write $x=\sqrt2+\sqrt[3]2$. Then $x-\sqrt2=\sqrt[3]2$ and so $$(x-\sqrt2)^3=x^3-3\sqrt 2x^2+6x-2\sqrt2=2.$$ A bit of rearrangement gives $$\sqrt2=\frac{x^3+6x-2}{3x^2+2}\in\Bbb Q(x).$$ It's clear then that also $\sqrt[3]2\in\Bbb Q(x)$.

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