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. 

Clearly $LHS\subseteq RHS$. Now it suffices to write down the minimal polynomial of $\sqrt{5}+\sqrt[3]{2}$ and note it has degree $6$.
