How to show field extension equality I've seen similar field extension questions on SE, but nothing with a third root, and I'm having trouble adapting any of those solutions to this problem.

So I'm trying to prove that $\mathbb{Q}(\sqrt{2}+{5}^{1/3})=\mathbb{Q}(\sqrt{2},{5}^{1/3})$. 

Now, $\mathbb{Q}(\sqrt{2},{5}^{1/3})$ contains both $\sqrt{2}$ and ${5}^{1/3}$, so $\mathbb{Q}(\sqrt{2}+{5}^{1/3})\subseteq\mathbb{Q}(\sqrt{2},{5}^{1/3})$, right? 

How then do I go about proving the reverse, that $\mathbb{Q}(\sqrt{2},{5}^{1/3})\subseteq\mathbb{Q}(\sqrt{2}+{5}^{1/3})$? I not even sure where to start. Any advice, tips?
An idea

$$r=\sqrt2+\sqrt[3]5\implies r^3-3\sqrt2\,r^2+6r-2\sqrt2=5\implies$$

$$(r^3+6r-5)^2=\left[(3r^2+2)\sqrt2\right]^2$$

The above shows that $\;\dim_{\Bbb Q}\Bbb Q(\sqrt 2+\sqrt[3]5)= 6\;$ (why? Show the polynomial $\;(r^3+6r-5)^2-\left[(3r^2+2)\sqrt2\right]^2\in\Bbb Q[x]\;$ is irreducible ), but since $\;\dim_{\Bbb Q}\Bbb Q(\sqrt2\,,\,\sqrt[3]5)=6\;$ (why?) and we already know $\;\Bbb Q(\sqrt2+\sqrt[3]5)\le\Bbb Q(\sqrt2\,,\,\sqrt[3]5)\;$ we get equality.
Actually I found an even simpler way, after receiving some hints from my professor. 

Let $\alpha=\sqrt{2} + \sqrt[3]{5}$, and notice that $(\alpha^2-2)/2\alpha=\sqrt[3]{5}$. Thus both $\sqrt{2}$ and $\sqrt[3]{5}$ are in $Q(\sqrt{2}+\sqrt[3]{5})$, and the two field extensions must be equal.