Is $\mathbb{Q}(\sqrt [n]{3})/\mathbb{Q}$ a splitting field of some polynomial

Is it true that the extension $\mathbb{Q}(\sqrt [n]{3})/\mathbb{Q}$ is the splitting field of some polynomial over $\mathbb{Q}$? My guess is no. But I can not prove it. Some observations I made are as $[\mathbb{Q}(\sqrt [n]{3}):\mathbb{Q}]=n$ any element of $\mathbb{Q}(\sqrt [n]{3})$ will satisfy a polynomial (irreducible over $\mathbb{Q}$) of degree atmost $n$. Splitting field of $x^{n}-3$ is $\mathbb{Q}(\sqrt [n]{3},\zeta_{n})$ where $\zeta_{n}$ is a primitive $n$th root of unity. I don't know whether these observations are of any use.

Any idea/help is most welcome. Thanks. Can this problem be solved without using Galois theory?

It is not the splitting field since it is not Galois. Such fields contain all the conjugate roots, $\zeta_n\sqrt[n]{3}$
Hint: If $K$ is a splitting field of some polynomial over $F$, and if some other irreducible polynomial over $F$ has a zero in $K$, then this other polynomial splits over $K$.