You were on the right track.

Starting in the same way as you did, let \begin{align*} k&=\mathbb{Q}\\[4pt] K&=\mathbb{Q}(\sqrt[3]2)\\[4pt] \end{align*} and let $\sigma_1,\sigma_2$ be embeddings of $K$ into $\overline{k}$ such that \begin{align*} \sigma_1(\sqrt[3]2)&=\sqrt[3]2\\[4pt] \sigma_2(\sqrt[3]2)&=\omega\sqrt[3]2\\[4pt] \end{align*} where $$ \omega = e^{\frac{2\pi i}{3}} = \frac{-1+i\sqrt{3}}{2} $$ Now let $F=\mathbb{Q}(\sqrt[3]2,\sqrt{3})$, and let $F_1,F_2$ denote $F\sigma_1(K),F\sigma_2(K)$, respectively.

Then $i\in F_2$, so $F_2$ has an element satisfying $x^2+1=0$, whereas $F_1$ has no such element.

It follows that $F_1,F_2$ are not isomorphic.

You were on the right track.

Starting in the same way as you did, let \begin{align*} k&=\mathbb{Q}\\[4pt] K&=\mathbb{Q}(\sqrt[3]2)\\[4pt] \end{align*} and let $\sigma_1,\sigma_2$ be embeddings of $K$ into $\overline{k}$ such that \begin{align*} \sigma_1(\sqrt[3]2)&=\sqrt[3]2\\[4pt] \sigma_2(\sqrt[3]2)&=\omega\sqrt[3]2\\[4pt] \end{align*} where $ \omega = e^{\large{\frac{2\pi i}{3}}} $.

Now

Then $\omega\in F_2$, so $F_2$ has an element satisfying $x^2+x+1=0$, whereas $F_1$ has no such element.

It follows that $F_1,F_2$ are not isomorphic.

You were on the right track.

Starting in the same way as you did, let \begin{align*} k&=\mathbb{Q}\\[4pt] K&=\mathbb{Q}(\sqrt[3]2)\\[4pt] \end{align*} and let $\sigma_1,\sigma_2$ be embeddings of $K$ into $\overline{k}$ such that \begin{align*} \sigma_1(\sqrt[3]2)&=\sqrt[3]2\\[4pt] \sigma_2(\sqrt[3]2)&=\omega\sqrt[3]2\\[4pt] \end{align*} where $$ \omega = e^{\frac{2\pi i}{3}} = \frac{-1+i\sqrt{3}}{2} $$ Now let $F=\mathbb{Q}(\sqrt{3})$, and let $F_1,F_2$ denote $F\sigma_1(K),F\sigma_2(K)$, respectively.

Then $i\in F_2$, so $F_2$ has an element satisfying $x^2+1=0$, whereas $F_1$ has no such element.

It follows that $F_1,F_2$ are not isomorphic.

You were on the right track.

Starting in the same way as you did, let \begin{align*} k&=\mathbb{Q}\\[4pt] K&=\mathbb{Q}(\sqrt[3]2)\\[4pt] \end{align*} and let $\sigma_1,\sigma_2$ be embeddings of $K$ into $\overline{k}$ such that \begin{align*} \sigma_1(\sqrt[3]2)&=\sqrt[3]2\\[4pt] \sigma_2(\sqrt[3]2)&=\omega\sqrt[3]2\\[4pt] \end{align*} where $$ \omega = e^{\frac{2\pi i}{3}} = \frac{-1+i\sqrt{3}}{2} $$ Now let $F=\mathbb{Q}(\sqrt[3]2,\sqrt{3})$, and let $F_1,F_2$ denote $F\sigma_1(K),F\sigma_2(K)$, respectively.

Then $i\in F_2$, so $F_2$ has an element satisfying $x^2+1=0$, whereas $F_1$ has no such element.

It follows that $F_1,F_2$ are not isomorphic.