The Galois cohomology group $H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z})$ classifies cyclic cubic extensions $K/\mathbb{Q}$ (specifically: the non-trivial elements correspond to Galois cubic field extensions $K/\mathbb{Q}$ together with a choice of isomorphism $\mathrm{Gal}(K/\mathbb{Q}) \cong \mathbb{Z}/3\mathbb{Z}$).


Let $k = \mathbb{Q}(\mu\_3)$. There are restriction and corestriction maps
$$\mathrm{Res}: H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z}) \to H^1(k, \mathbb{Z}/3\mathbb{Z}), \quad \mathrm{Cores}: H^1(k, \mathbb{Z}/3\mathbb{Z}) \to H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z}).$$ 
Restriction followed by corestriction is multiplication by $2$ on $H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z})$. As each element is $3$-torsion, it follows that $\mathrm{Res}$ is injective and that $\mathrm{Cores}$ is surjective.


But as $\mu\_3 \subset k$, it follows from Kummer theory that 
$$H^1(k, \mathbb{Z}/3\mathbb{Z}) \cong H^1(k, \mu\_3) \cong k^\*/k^{\*3}.$$
Composing with corestriction, we therefore obtain a surjective map
$$f: k^{\*}/k^{\*3} \to H^1(\mathbb{Q}, \mathbb{Z}/3\mathbb{Z}).$$



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I know that the corestriction $H^1(k, \mu\_3) \cong k^\*/k^{3\*} \to \mathbb{Q}^\*/\mathbb{Q}^{\*3} \cong H^1(\mathbb{Q}, \mu\_3)$ is just usual norm map. But this doesn't seem to help here. 



Cyclic cubic extensions and Kummer theory
It's just the map 


$$x \mapsto y = \frac{x}{x^{\sigma}},$$


where the corresponding degree three extension of $\mathbb{Q}$ is the degree three subfield of $k(y^{1/3})$. The point is that it is obvious from the restriction map that


$$H^1(\mathbb{Q},\mathbb{Z}/3 \mathbb{Z})
= (k^{\times}/k^{\times 3})^{G = {\chi}},$$


where $\chi$ is the non-trivial character of $G = \mathrm{Gal}(\mathbb{Q}(\zeta\_3)/\mathbb{Q}) = \mathbb{Z}/2 \mathbb{Z}$.


(**Added** Here $M^{\chi}$ means what is says on the tin. If $\sigma \in G$ and $m \in M^{\chi}$, then $\sigma m = \chi(\sigma) m$.)


And basic Kummer theory also says that degree 3 cyclic extensions $K$ of $\mathbb{Q}$ have the form $K(\zeta\_3) = \mathbb{Q}(\zeta\_3)(\alpha^{1/3})$ where


$$\sigma \alpha = \alpha^{-1} \mod k^{\times}/k^{\times 3}.$$


The same basic structure holds *mutatis mutandis* with $\mathbb{Q}$ replaced by any number field $F$, and $3$ replaced by $p$, and $G = \chi$ where now $\chi$ is the mod-p cyclotomic character of $G = \mathrm{Gal}(F(\zeta\_p)/F)$, which is the canonical (possibly trivial) map $G \rightarrow (\mathbb{Z}/p \mathbb{Z})^{\times}$. And now the map from $k^{\times}/k^{\times p}$ is just the projection to the $\chi$-eigenspace.


**Added**: If you want an explicit polynomial, you can, of course, use Galois theory to do so. In fact, everything in this question one can (and I do) teach in the introductory undergraduate Galois theory course. To spell out the elementary details, you want an element of $k(y^{1/3})$ which is fixed by the order two element $\sigma \in \mathrm{Gal}(k(y^{1/3})/\mathbb{Q})$ (there is an obvious splitting from $\mathrm{Gal}(k/\mathbb{Q}) \rightarrow \mathrm{Gal}(k(y^{1/3}/\mathbb{Q})$). The obvious element to take is thus


$$z = y^{1/3} + \sigma y^{1/3} = y^{1/3} + y^{-1/3},$$


which is a root of


$$T^3 - 3 T - (y + y^{-1}) = T^3 - 3 T - \left(\frac{x}{x^{\sigma}} + \frac{x^{\sigma}}{x}\right) 
= T^3 - 3 T - \frac{Tr(x^2)}{N(x)} \in \mathbb{Q}[T].$$


