Find an element $\theta \in \mathbb{R}$ such that $\mathbb{Q}(\sqrt{2},\sqrt[3]{5}) = \mathbb{Q}(\theta)$ 
Find an element $\theta \in \mathbb{R}$ such that
  $\mathbb{Q}(\sqrt{2},\sqrt[3]{5}) = \mathbb{Q}(\theta)$

I must find one element $\theta$ that
$$\mathbb{Q}(\sqrt{2},\sqrt[3]{5}) \subseteq \mathbb{Q}(\theta)$$
and
$$\mathbb{Q}(\theta)\subseteq \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$$
For example, I know that $\mathbb{Q}(\sqrt{2},\sqrt[3]{5})\subseteq \mathbb{Q}(\sqrt{2}+\sqrt[3]{5})$, because the sum of these two elements must also be in the set. I don't know, however, if the inverse inclusion holds. For example, I must be able to form $\sqrt{2}$ and $\sqrt[3]{5}$ using $\sqrt{2}+\sqrt[3]{5}$ using only multiplications of $\sqrt{2}+\sqrt[3]{5}$ by itself, using its inverse, and summing with the results of the multiplications, right?
$(\sqrt{2}+\sqrt[3]{5})^2 = 2 + 2\sqrt{2}\sqrt[3]{5} + \sqrt[3]{5^2}$ which isn't helpful at all, so I think this is not the right candidate. 

Take $\theta=\sqrt{2}{\sqrt[3]{5}}$
We have that $(\sqrt{2}\sqrt[3]{5})^2=2\sqrt[3]{5^2} \Rightarrow \sqrt[3]{5^2} \in \mathbb{Q}(\theta)$
Therefore $\sqrt{2}=\frac{(\sqrt{2}\sqrt[3]{5})\sqrt[3]{5^2}}{5} \in \mathbb{Q}(\theta) \Rightarrow \sqrt{2}\in \mathbb{Q}(\theta)$
Now with the same way we can prove that $\sqrt[3]{5} \in \mathbb{Q}(\theta)$ by noticing that $\sqrt[3]{5}=\frac{\sqrt{2}(\sqrt{2}\sqrt[3]{5})}{2}$
Thus $\mathbb{Q}(\sqrt{2},\sqrt[3]{5}) \subseteq \mathbb{Q}(\theta)$
Also it is obvious that $\mathbb{Q}(\theta) \subseteq \mathbb{Q}(\sqrt{2},\sqrt[3]{5})$
So you have the result.
