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!
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 $\alpha=\frac{3-6\gamma-\gamma^3}{2-3\gamma^2}\in K$. Therefore $\beta=\gamma-\alpha\in K$ and so $K=L$.
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 $\sqrt2+\sqrt[3]3$.