Question about a degree 6 Galois extension $K(\omega) : \mathbb{Q}$ 
Let $K : \mathbb{Q}$ be a finite Galois extension with Galois group $\mathbb{Z}/3$
Let $K(\omega) : \mathbb{Q}$ be a Galois extension with Galois group $G \cong \mathbb{Z}/6 \cong \mathbb{Z}/3 \times \mathbb{Z}/2$, where $\omega $ is the standard cube root of unity (and $\omega \notin K$).
Let $\tau,\sigma \in G$ be such that the fixed field of <$\tau$> is $K$ and the fixed field of <$\sigma$> is $\mathbb{Q}(\omega)$.
Show that there exists an element $\beta \in K(\omega)$ such that both of the following hold:
(1) $\beta \notin \mathbb{Q}(\omega)$ and $\beta^3 \in \mathbb{Q}(\omega)$
(2)$\beta + \tau(\beta) \in K$

I have tried finding such an element using Cardano's method on the cubic polynomial with one of its roots $\beta + \tau(\beta)$ and other two roots $\omega \beta + \omega^2 \tau(\beta)$ and  $\omega^2 \beta + \omega \tau(\beta)$ but that didn't work out. I'm not sure what else to do. I've been stuck on this for a while.
 A: Thanks to @Jyrki Lahtonen in the comments I think I understand what I missing. I will write out what I believe to be the solution. I also included the larger problem this was a part of for context. 
First notice that $\sigma^3 = \tau^2 = id_{K(\omega)}$
We know that $id_{K(\omega)}$, $\sigma$, and $\sigma^2$ are linearly independent over $K(\omega)$ and so by the linear independence of field homomorphisms, we have that
$id + \omega \sigma + \omega^2 \sigma^2 \neq 0$
So there exists some $x \in K(\omega)$ such that 
$x + \omega \sigma (x) + \omega^2 \sigma^2 (x) \neq 0$
Denote the expression on the LHS as $\beta$.
(1) To see that $\beta$ is not in $\mathbb{Q}(\omega)$, notice that $\sigma (\beta) = \sigma (x) + \omega \sigma^2 (x) + \omega^2 x = \omega^2 \beta \neq \beta$ and so $\beta \notin \mathbb{Q}(\omega)$
Furthermore, by computing $\beta^3$ we get the (not very elegant) equation:
$\beta^3 = x^3 + \sigma(x^3) + \sigma^2(x^3) + 3 (\omega x^2 \sigma(x) + \omega^2 x^2 \sigma^2 (x) + \omega \sigma (x^2) \sigma^2 (x) + \omega^2 x \sigma (x^2) + \omega x \sigma^2 (x^2) + \omega^2 \sigma^2(x) \sigma (x)) + 6 x \sigma (x) \sigma^2(x) $
And we can see that $\sigma(\beta^3) = \beta^3$. Thus $\beta^3 \in \mathbb{Q}(\omega)$
(2) This is simple. $\tau (\beta + \tau(\beta)) = \tau (\beta) + \tau^2(\beta) = \beta + \tau(\beta)$ and we are done.
To add some context, this was a larger part of a problem to show that with the conditions as given in the problem statement, there exists $\beta \in K(\omega)$ such that (1) and (2) hold in addition to 
(3) There exists an irreducible depressed cubic $X^3 + pX + q \in Irred(\mathbb{Q})$ with $\alpha = \beta + \tau(\beta)$ as a root with splitting field $K$ and 
(4) $\Delta := -27q^2 - 4p^3$ being equal to the square of some element of $\mathbb{Q}$
The cubic equation turns out to be $X^3 - 3\beta \tau(\beta)X - (\beta^3 + \tau(\beta^3))$. You can see through any discussion of Cardano's method by taking your roots to be 
$\beta + \tau(\beta)$
$\omega^2 \beta + \omega \tau(\beta)$
$\omega \beta + \omega^2 \tau(\beta)$
It is a polynomial over $\mathbb{Q}$ because the coefficients are fixed by both $\sigma$ and $\tau$ (which generate the Galois group of $K(\omega) : \mathbb{Q}$). This can be checked by similar (and less tedious) calculations as before. The polynomial is irreducible because it none of its roots are in $\mathbb{Q}$ (Since $\beta + \tau(\beta) \in K$ is not fixed by $\sigma$ and so none of the roots are). 
Property (4) is true because the Galois group of the polynomial (i.e. the Galois group of the extension of the splitting field of the polynomial $K : \mathbb{Q}$) is isomorphic to the alternating group $A_3$. 
