I'm trying to show that $\mathbb{Q} [x ] / ( x^3-2)$ is not isomorphic to $\mathbb{Q} [x ] / ( x^3-3)$.
My attempt : Since $\mathbb{Q} [x ] / ( x^3-2)$ is isomorphic to $\mathbb{Q} [2^{1/3}]$ and $\mathbb{Q} [x ] / ( x^3-3)$ is isomorphic to $\mathbb{Q} [3^{1/3}]$, I will show that $\mathbb{Q} [2^{1/3}]$ is not isomorphic to $\mathbb{Q} [3^{1/3}].$ Is it correct idea?
Suppose that $$ f:\mathbb{Q}[2^{1/3}] \rightarrow \mathbb{Q}[3^{1/3}]$$ is an isomorphism.
Let $\alpha\in\mathbb{Q}[2^{1/3}]$ be such that $f(\alpha)=3^{1/3}$. Then $f(\alpha^3)=3$. This gives $\alpha^3 = 3$ in $\mathbb{Q}[2^{1/3}]$.
To see that this is impossible, write $\alpha = a \cdot 1 + b \cdot 2^{1/3} + c \cdot 2^{2/3}$ with rational $a, b, c$. Using the traces $\mathrm{Tr}_{\mathbb{Q}}^{\mathbb{Q}[2^{1/3}]} (\alpha)$, $\mathrm{Tr}_{\mathbb{Q}}^{\mathbb{Q}[2^{1/3}]}(2^{1/3} \alpha)$, and $\mathrm{Tr}_{\mathbb{Q}}^{\mathbb{Q}[2^{1/3}]}(2^{2/3} \alpha)$, we obtain that $$ 0 = 3a, \ \ 0=6c, \ \ 0=6b. $$ This is a contradiction since $a=b=c=0$ makes $\alpha=0$ and $\alpha^3=3$.
We already know at MSE that $[\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3}):\mathbb{Q}]=9$, see here. It says that $\sqrt[3]{3}\not\in \mathbb{Q}(\sqrt[3]{2})$. So $\alpha^3=3$ cannot hold in $\mathbb{Q}(\sqrt[3]{2})$, so that $\mathbb{Q}(\sqrt[3]{2})$ is not field isomorphic to $\mathbb{Q}(\sqrt[3]{3})$.
