I'm working on the number field $\mathbb{Q}(\sqrt{-3})$, and I want to find elements $\alpha \in \mathbb{Q}(\sqrt{-3})$ such that the polynomial $X^3-\frac{\overline{\alpha}}{\alpha^2}$ be irreducible. In other words I need $\frac{\overline{\alpha}}{\alpha^2}$ to be a cube, but: $$\frac{\overline{\alpha}}{\alpha^2}=\frac{\overline{\alpha}^3}{N_{\mathbb{Q}(\sqrt{-3})/\mathbb{Q}}(\alpha)}$$ And thus, we just need the norm $N_{\mathbb{Q}(\sqrt{-3})/\mathbb{Q}}(\alpha)$ to be a cube. If $\alpha=a+b\sqrt{3}$, then the norm is equal $a^2+3b^2$. So I want to find solutions to equations:
$$a^2+3b^2=k^3, \ a,b,k \in \mathbb{Q}$$
Does this kind of equation have a way to find solutions of it?
Thanks a lot.