Question is to prove that $\mathbb{Q}(\sqrt[3]{2})$ is not in any cyclotomic extension of $\mathbb{Q}$.
As i was not so sure how to proceed i did the following thing(which may possibly be not so relevant to the Question but hope it would give me some hint to proceed). As $\mathbb{Q}(\sqrt[3]{2})$ is not Galois over $\mathbb{Q}$, I tried Computing Splitting Field and corresponding galois Group and ended up in concluding that Galois Group is isomorphic to $S_3$. As $S_3$ is not abelian Group there could possibly no Sub field $K$ of Cyclotomic field with $Gal(\frac{K}{\mathbb{Q}})\cong S_3$.
I am helpless after this.
Please let me know am i going in a correct path?? Any suggestion/hint would be appreciated :) Thank You
You have already solved it. I would phrase it as follows: Normal subextensions of abelian extensions are also abelian. Therefore, $\mathbb{Q}(\sqrt[3]{2})$ is not contained in any abelian extension, since then also its normal closure $\mathbb{Q}(\zeta_3,\sqrt[3]{2})$ would be contained, but this has Galois group $S_3$.