# Showing that $\mathbb{Q}(\sqrt{2}) \subseteq \mathbb{Q}(\sqrt{2}+\sqrt[3]{2})$

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!

• 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 $\alpha, \alpha^2$. – WimC Sep 2 at 18:52
• – Jyrki Lahtonen Sep 2 at 18:59

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 $$\sqrt[3]2\in\Bbb Q(x)$$.