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

Question

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!

Answer 1

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

Answer 2

Let $F=\mathbb{Q}(\sqrt{2}+\sqrt[3]{2})$ and $L=\mathbb{Q}(\sqrt{2},\sqrt[3]{2})$. Clearly, $F\subseteq L$. Assume that $\sqrt{2},\sqrt[3]{2} \notin F$. Now, note that $F(\sqrt{2})=F(\sqrt[3]{2})=L$.
The minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}$ is $x^2-2$. So, $m_{\sqrt{2},F}|x^2-2$ in $F[x]$ and it should be equal to $x^2-2$ by our assumption. Thus we get $[L:F]=2$.
From this we get $deg (m_{\sqrt[3]{2},F})=2$ and $m_{\sqrt[3]{2},F}|x^3-2$ in $F[x]$ as it is minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}$. From this we get $m_{\sqrt[3]{2},F}=(x-\sqrt[3]{2})(x-\sqrt[3]{2}\omega)$ or $m_{\sqrt[3]{2},F}=(x-\sqrt[3]{2})(x-\sqrt[3]{2}\omega^2)$. But this isn't possible because after multiplying these factors we will not get coefficients from $F$. Thus we get contradiction. So, $\sqrt{2},\sqrt[3]{2} \in F$, which implies $F=L$.