Prove that $\mathbb{Q}(\sqrt{2},\sqrt[3]{3})=\mathbb{Q}(\sqrt{2}\sqrt[3]{3})=\mathbb{Q}(\sqrt{2}+\sqrt[3]{3})$ Showing that $\mathbb{Q}(\sqrt{2},\sqrt[3]{3})$ has degree 6 over $\mathbb{Q}$ is straighforward:
It contains $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{3})$ which are degree 2 and 3 over $\mathbb{Q}$ and since 2 and 3 are coprime it follows that $[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}]=6$.
Then since $\mathbb{Q}(\sqrt{2}\sqrt[3]{3})\subset \mathbb{Q}(\sqrt{2},\sqrt[3]{3})$ and $x^6-72$ is the minimal polynomial of $\sqrt{2}\sqrt[3]{3}$ it follows that  $[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}(\sqrt{2}\sqrt[3]{3})]=1$ and so $\mathbb{Q}(\sqrt{2},\sqrt[3]{3})=\mathbb{Q}(\sqrt{2}\sqrt[3]{3})$.
But I can't figure out how to prove the last equality. Any help is appreciated!
 A: So after some thinking, I came up with two solutions:
Solution 1
Let $\alpha=\sqrt{2},\beta=\sqrt[3]{3}$ and $\gamma=\alpha+\beta$. Let $L=\mathbb{Q}(\alpha, \beta)$, $K=\mathbb{Q}(\gamma)$ and suppose that $\alpha,\beta\not\in K$. Since if one of $\alpha,\beta$ is in $K$, we get the other one for free it follows that $L=K(\alpha)=K(\beta)$. Then the minimal polynomial of $\alpha$ in $K[X]$ is of degree 2 since $\alpha$ is a root of $X^2-2$ and we assumed $\alpha\not\in K$. Since $X^3-3$ has one real root and two non-real roots it follows that it is the minimal polynomial of $\beta$ in $K[X]$. Hence we conclude
$$2=[K(\alpha):K]=[L:K]=[K(\beta):K]=3.$$
Which is a contradiction and so $L=K$ as desired.
Solution 2
We know that $\gamma-\alpha=\beta$ so $(\gamma-\alpha)^3=3$. Expanding the left side we get
$$\gamma^3-3\gamma^2\alpha+6\gamma-\alpha^3=3$$
and so $\alpha=\frac{\gamma^3+6\gamma-3}{2+3\gamma^2}\in K$. Therefore $\beta=\gamma-\alpha\in K$ and so $K=L$.
A: This is an old question but here is another method.
The splitting field of $(x^{3}-3)(x^{2}-2)$ is $K=\Bbb{Q}(\sqrt[3]{3},\omega,\sqrt{2})$ and it is the compositum of the splitting fields $\Bbb{Q}(\sqrt[3]{3},\omega)$ and $\Bbb{Q}(\sqrt{2})$. Here $\omega$ denotes the  primitive third root of unity.
And $\text{Gal}(\Bbb{Q}(\sqrt[3]{3},\omega)/\Bbb{Q})\cong S_{3}$  ,
$\text{Gal}(\Bbb{Q}(\sqrt{2})/\Bbb{Q})\cong\Bbb{Z_{2}}$ . and $\text{Gal}(K/\Bbb{Q})\cong D_{12}$
If we determine the action of the automorphisms $f_{1},f_{2},f_{3}$ by:-
$$f_{1}=\begin{cases} \sqrt{2}\mapsto -\sqrt{2}\\ \sqrt[3]{3}\mapsto \sqrt[3]{3}\\ \omega \mapsto \omega\end{cases}$$
$$f_{2}=\begin{cases} \sqrt{2}\mapsto \sqrt{2}\\ \sqrt[3]{3}\mapsto \omega \sqrt[3]{3}\\ \omega \mapsto \omega\end{cases}$$
$$f_{3}=\begin{cases} \sqrt{2}\mapsto \sqrt{2}\\ \sqrt[3]{3}\mapsto  \sqrt[3]{3}\\ \omega \mapsto \omega^{2}\end{cases}$$ . Then it is easy to see that $\text{Gal}(K/\Bbb{Q})$ is generated by $\{f_{1},f_{2},f_{3}\}$ and only $f_{3}$ fixes $\sqrt{2}+\sqrt[3]{3}$.
Hence $\sqrt{2}+\sqrt[3]{3}$ is fixed by $\langle f_{3}\rangle$ and hence the fixed field of $\langle f_{3}\rangle$ is $\Bbb{Q}(\sqrt{2}+\sqrt[3]{3})$
By Fundamental Theorem of Galois theory $[\Bbb{Q}(\sqrt{2}+\sqrt[3]{3}):\Bbb{Q}]=\text{index of the subgroup} \langle f_{3}\rangle = \frac{12}{2}=6$
Note that this method can be directly applied to all your cases and shown that all of them are indeed the fixed field of $\langle f_{3}\rangle$ and hence must be equal by Fundamental Theorem.
A: Here's a solution which hardly contains any kind of calculation whatsoever.
Write $K = \mathbb{Q}(\sqrt{2}+\sqrt[3]{3})$ and $L = \mathbb{Q}(\sqrt{2},\sqrt[3]{3})$. Assume $K \neq L$: we will derive a contradiction.
It follows from the tower law that $[K:\mathbb{Q}]$ equals $1$, $2$ or $3$. However $[K:\mathbb{Q}]=1$ and $[K:\mathbb{Q}]=2$ are impossible, since adjoining the element $\sqrt{2}$ of degree $2$ would produce the full field $L$ (since it contains both $\sqrt{2}$ and $\sqrt{2}+\sqrt[3]{3}$). So it must be the case that $[K:\mathbb{Q}]=3$.
Now consider $K(\sqrt[3]{3})$, which again equals $L$. Since $[L:K]=2$, the element $\sqrt[3]{3}$ must satisfy a quadratic equation over $K$. But that means that $X^3-3$ splits into a linear and a quadratic factor over $K$, which would imply $[L:K]=1$, giving the desired contradiction.
A: Show that $x^3-3$ is irreducible over $\Bbb{Q}$ and $x^2-2$ is irreducible over $\Bbb{Q}(\sqrt[3]3)$,
which implies that $[\Bbb{Q}(\sqrt2,\sqrt[3]3):\Bbb{Q}]=6$.
Next show that $\sqrt2\sqrt[3]3$ has $6$ distinct $\Bbb{Q}$-conjugates, which implies that $[\Bbb{Q}(\sqrt2\sqrt[3]3):\Bbb{Q}]=6$.
Same for $\sqrt2+\sqrt[3]3$.
