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. 
 A: 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. 
A: Clearly $LHS\subseteq RHS$. Now it suffices to write down the minimal polynomial of $\sqrt{5}+\sqrt[3]{2}$ and note it has degree $6$.
A: Let $F=\mathbb{Q}(\sqrt{5}+\sqrt[3]{2})$ and $L=\mathbb{Q}(\sqrt{5},\sqrt[3]{2})$. Clearly, $F\subseteq L$. Assume that $\sqrt{5},\sqrt[3]{2} \notin F$. Now, note that $F(\sqrt{5})=F(\sqrt[3]{2})=L$.
The minimal polynomial of $\sqrt{5}$ over $\mathbb{Q}$ is $x^2-5$. So, $m_{\sqrt{5},F}|x^2-5$ in $F[x]$ and it should be equal to $x^2-5$ 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, $\sqrt{5},\sqrt[3]{2} \in F$, which implies $F=L$.
