I have problems with this exercise:
I need to show that $\Bbb Q(\sqrt{2}+\sqrt[3]{3})$ is normal over $\mathbb{Q}$.
My idea was to find the minimal polynomial of $\sqrt{2}+\sqrt[3]{3}$ and show that any root is in $\Bbb Q(\sqrt{2}+\sqrt[3]{3})$. The problem is that I cannot find it.
Here are some calculations that I have done:
$\sqrt{2}+\sqrt[3]{3}=a \\ a-\sqrt[3]{3}=\sqrt{2} \\ (a-\sqrt[3]{3})^2=2 \\ a^2-2a\sqrt[3]{3}+\sqrt[3]{3^2}-2=0 \\ a^2-a\sqrt[3]{3}-a\sqrt[3]{3}+\sqrt[3]{3^2}-2=0 \\ a^2-a\sqrt[3]{3} -\sqrt[3]{3}(a-\sqrt[3]{3})-2=0 \\ a^2-a\sqrt[3]{3} -\sqrt[3]{3}\sqrt{2}-2=0 \\a^2-\sqrt[3]{3}(a-\sqrt{2})-2=0 \\a^2-\sqrt[3]{3}(\sqrt[3]{3})-2=0 \\a^2-2=\sqrt[3]{3^2} \\(a^2-2)^3-9=0 \\$
Let $a=\sqrt{2} + \sqrt[3]{3}$. Then $3=(a-\sqrt{2})^3$. Expanding everything out you will see that $\sqrt{2}$ can be represented as some rational function in $a$, which indicates the field $\mathbb{Q}(\sqrt{2}+ \sqrt[3]{3})$ is the field $\mathbb{Q}(\sqrt{2}, \sqrt[3]{3})$.
Assume that $\mathbb{Q}(\sqrt{2}+ \sqrt[3]{3})$ is normal over $\mathbb{Q}$, then it must be normal over subfield $\mathbb{Q}(\sqrt{2})$. Using the tower rule should give you $$[\mathbb{Q}(\sqrt{2}+ \sqrt[3]{3}):\mathbb{Q}(\sqrt{2})]=3,$$ which show that $x^3-3$ is the minimal polynomial of $\sqrt[3]{3}$ over $\mathbb{Q}(\sqrt{2})$. But only one root of $x^3-3$ lies inside $\mathbb{Q}(\sqrt{2}+ \sqrt[3]{3})$, which contradicts the normality assumption.