Show that $\mathbb{Q}(\sqrt{5}+\sqrt[3]{2})=\mathbb{Q}(\sqrt{5},\sqrt[3]{2})$. 
Show that $\mathbb{Q}(\sqrt{5}+\sqrt[3]{2})=\mathbb{Q}(\sqrt{5},\sqrt[3]{2})$.

I've got that $\big[\mathbb{Q}(\sqrt{5}+\sqrt[3]{2}):\mathbb{Q}\big] \in \{1,2,3,6\}$ because it's going to divide $\big[\mathbb{Q}(\sqrt{5},\sqrt[3]{2}):\mathbb{Q}\big]=6$. Clearly it is not $1$. I want to show that it is not $2$ or $3$. So I'm saying that if it is $2$ then $\alpha = \sqrt{5}+\sqrt[3]{2}$ satisfies the relation
$$\alpha^2 + b\alpha = k$$ for $b,k \in \mathbb{Q}$ since $\alpha$ will be the root of a monic irreducible polynomial of degree  $2$. How can I obtain a contradiction from this? I also need to do the degree $3$ case somehow. 
Also if there is a better way to do this than what I'm doing I'd be excited to learn about it. 

Since you know that $[\mathbb Q[\sqrt[3]{2},\sqrt 5]:\mathbb Q] = 6$, you know that each of the six values $$1,\sqrt{5},\\\sqrt[3]{2},\sqrt[3]{2}\sqrt{5},\\\sqrt[3]{4},\sqrt[3]{4}\sqrt 5\tag{1}$$ 
are linearly independent over $\mathbb Q$.
Now $$(\sqrt[3]2+\sqrt 5)^2=5\cdot 1 + 2\cdot \sqrt[3]2\sqrt 5 + 1\cdot\sqrt[3]4$$
Is it possible for $1,\sqrt[3]2+\sqrt5,(\sqrt[3]2+\sqrt 5)^2$ to be linearly dependent over $\mathbb Q$? 
Do the same with by adding the cube $(\sqrt[3]2+\sqrt5)^3$.
Another way to look at it use (1) as a basis, and write elements of the field as:
$$(a,b,c,d,e,f)\to a\cdot 1 + b\cdot \sqrt5+c\sqrt[3]2+d\sqrt[3]2\sqrt5+e\sqrt[3]4+f\sqrt[3]4\sqrt5$$
Then $$\begin{align}(1,0,0,0,0,0)&\leftrightarrow 1\\(0,1,1,0,0,0)&\leftrightarrow \sqrt 5+ \sqrt[3]2\\(5,0,0,2,1,0)&\leftrightarrow (\sqrt5+\sqrt[3]2)^2\\
(2,5,15,0,0,3)&\leftrightarrow (\sqrt5+\sqrt[3]2)^3
\end{align}$$
And those four vectors are "obviously" linearly independent. 
