Can $\sqrt{3}$ be written as a polynomial expression in $\sqrt[3]{3}$ and $\zeta_3$ I believe that it cannot be done. But now I can only think of the method using the basis of $\Bbb Q(\zeta_3,\sqrt[3]{3})$, which is brute force and tedious. So I am now searching for a rather simple method. Could someone please tell me if there is some simple method? Thanks!
 A: No. The field $K=\Bbb Q(\zeta_3,\sqrt[3]3)$ has degree $6$ over $\Bbb Q$
and Galois group $S_3$. It has a unique quadratic subfield, $\Bbb Q(\zeta_3)=\Bbb Q(\sqrt{-3})$.
A: Galois theory provides a roundabout way of doing this.  Consider the following lemma:
Lemma: If $K/F$ is a Galois extension, then the minimal polynomial in $F[x]$ for any $a \in K$ has as its roots the elements in the orbit of $a$ under the action of $\text{Gal}(K/F)$.  That is, if $S = \{ \phi(a) \ | \ \phi \in \text{Gal}(K/F) \}$, then $\displaystyle \min_a(x) = \prod_{u_k \in S} (x-u_k)$.
Proof: This is a consequence of the fact that a polynomial is irreducible $\iff$ its Galois group acts transitively on its roots.  For a proof of this fact, see Theorem 2.9(b) here.  Notice that $\displaystyle \min_a(x)$ will be an irreducible polynomial (by definition), and its Galois group will be a subgroup of $\text{Gal}(K/F)$.  This latter fact is because, if $L \subseteq K$ is the splitting field of $\displaystyle \min_a(x)$, every $F$-automorphism of $L$ extends to an $F$-automorphism of $K$. $\qquad \blacksquare$

Let $K = \mathbb{Q}(\sqrt[3]{3}, \zeta_3)$.  Note $K/\mathbb{Q}$ is a Galois extension since finite extensions of $\mathbb{Q}$ are separable, and it is the splitting field for $f(x) = x^3 - 3$.  Now assume $\sqrt{3} \in K$.  Because we know the minimal polynomial of $\sqrt{3}$ to be quadratic, the orbit of $\sqrt{3}$ under the action of $\text{Gal}(K/F)$ should have two elements per the lemma.What is $\text{Gal}(K/F)$?  It is a $6$-element group $\cong S_3$ generated by the following two automorphisms: complex conjugation and $\sigma$ defined where $\sqrt[3]{3} \mapsto \zeta_3 \sqrt[3]{3}$.  
Clearly complex conjugation fixes $\sqrt{3}$. If $\sigma$ does not fix $\sqrt{3}$, it is necessarily sent to $-\sqrt{3}$.  But $\sigma$ has order $3$, i.e. $\sigma^3 = \text{id}$, whereas we would have $\sigma^3(\sqrt{3}) = - \sqrt{3}$.  Not possible.  So $\sqrt{3}$ is fixed by every element of $\text{Gal}(K/F)$, which implies $g(x) = x - \sqrt{3}$ is its minimal polynomial over $\mathbb{Q}$.
This is a contradiction arising from our assumption that $\sqrt{3} \in K$.  We can conclude $\sqrt{3} \notin K$.
