Let $K$ be the splitting field of $P(X)=X^3 - 3X^2 + 3$ over $\mathbb{Q}$. Determine $G = \text{Gal}(K/\mathbb{Q})$. For every subgroup $H < G$, find $K^H$.
By Eisenstein's Criterion, $P(X)=X^3-3X^2+3$ is irreducible. $D(P)=-4\times (-81)\times 3-27\times 9=3^5$. Since $\text{Char}(\mathbb{Q})=0$, the Galois group $G$ is $\Sigma_3$. Therefore, $H$ is $\{e\}$ or $\{e, (12)\}$ or $\{e, (13)\}$ or $\{e, (23)\}$ or $A_3$ or $\Sigma_3$.
I can not find the radical solution of this equation so I can't proceed anymore. Any hints are appreciated, thanks in advance.
First let's determine the Galois group of $P(X) = X^3 - 3X^2 + 3$.
The discriminant of the general trinomial $X^3 + aX^2 + b$ is $\Delta = -b(4a^3 + 27b)$. In our case we have $\Delta = 81$ which is a square. This tells us that the Galois group is $A_3$.
The only subgroup of that is the trivial group, so there are no nontrivial subfields. The $A_3$ invariant subfield would just be $\mathbb Q$.