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. 
 A: For your candidate $\theta =\sqrt{2} + \sqrt[3]{5}$ we have
$$(\theta -\sqrt{2})^3=5\\
\theta^3 + 6\theta-(3 \theta^2 +2)\sqrt{2} =5\\
\sqrt{2}=\frac{\theta^3 + 6\theta - 5}{3\theta^2 +2}$$ and then use
$\sqrt[3]{5}= \theta -\sqrt{2}$
A: It is certainly true that your guess of $\theta=\sqrt2+\sqrt[3]5$ would work, and there are very many methods of showing this, though none is as quick as using instead the product of $\sqrt2$ and $\sqrt[3]5$ as suggested by @Marios.
For instance, consider the field $k=\Bbb Q(\sqrt2\,)$,whose ring of integers is $\Bbb Z[\sqrt2\,]$, known (or easily shown) to be a Principal Ideal Domain. In this ring, it happens that $5$ is still prime, though even if $5$ were to factor as $5=\pi\pi'$, my argument below would still apply:
Since $5$ is square-free in $\Bbb Z[\sqrt2\,]$, we then know that $\sqrt[3]5\notin k$, and as a result $\sqrt2+\sqrt[3]5\notin k$, and this quantity therefore generates the field $k(\sqrt[3]5\,)=\Bbb Q(\sqrt2,\sqrt[3]5\,)$.
A: 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.
