I attempted to show that for $K =\mathbb{Q}(i\sqrt{2}) \subset L=K(\sqrt[3]{6+i})$ Field Extension, that $L$ is not a normal Field Extension of $K$, but I am not sure whether this is the right way to go about it. I would really appreciate any feedback! My attempt:
First, since $i$ cannot be generated in $K$, $x^3-(6+i) \notin K[x]$. So I have to find a different minimal polynomial $P$ s.t $L=K[x]/(P)$. To do so, I tried to work backwords: $$x=\sqrt[3]{6+i} \implies x^3=6+i \implies x^3-6=i \implies (x^3-6)^2=-1 \implies (x^3-6)^2+1=0$$ Then $P=(x^3-6)^2+1=x^6-12x^3+37$ is a polynomial in $K[x]$ with $\sqrt[3]{6+i}$ as a root. We want P to be minimal, so I tried to show that it is in fact irreducible by looking at the roots of $x^3=6+i$ and $x^3=6-i$ (since $(x^3-6)^2+1=0 \implies x^3=6 \pm i$). Using De Moivre's theorem, we can find 3 complex roots for each cubic equation, so a total of 6 roots, which all are not in $K$, so P is minimal.
From here I argue that $\sqrt[3]{6-i} \notin L$, which is one of the roots, therefore $L$ is not a splitting field for $P$ and thus also not a normal Field Extension of $K$
Is this a valid argument?
Thanks!