How are these two discriminants of polynomials related? I was wondering the following problem, suppose we have $a,b$ such that $\mathbb Q(a)=\mathbb Q(b).$ Then consider the minimal polynomial of $a$, say $m_a(x)$ and $m_b(x)$ defined analogously. Is it possible that the discriminants of these two polynomials are related in some way? Mainly I am trying to solve a question and if disc $(m_a(x))=A^2$ disc $(m_b(x))$ for some $A\in \mathbb{Q}$ then that would complete my answer but I am struggling to see if this is true? (specifically I was considering the third degree extension of $\mathbb{Q}$)
Many thanks in advance!!
Edit:
This is the question I was trying to do. Whilst there were quite a few answers, I was trying to go about my own way by using the method I described above. In particular, I was trying to show the forward implication and so I assumed $\mathbb Q(a)=\mathbb Q(b)$, where the latter is meant to be radical. Then I found the latter has discriminant of $-27b^3$ and so if the assertion I had above was correct then the forward implication is shown.
 A: For a cubic extension of $\mathbb{Q}(a)$ of $\mathbb{Q}$, we have that the Galois closure is $\mathbb{Q}(a,\sqrt{\Delta_{a}})$. Since $\mathbb{Q}(a)=\mathbb{Q}(b)$, we have that their galois closures are the same and so $\mathbb{Q}(a,\sqrt{\Delta_{a}})=\mathbb{Q}(b,\sqrt{\Delta_{b}})$ with Galois Group at worst $S_{3}$. Therefore, by the fundamental theorem of Galois Theory, we have $\mathbb{Q}(\sqrt{\Delta_{a}})=\mathbb{Q}(\sqrt{\Delta_{b}})$ (fixed field of $C_{3} \subseteq S_{3}$), and so $\Delta_{a}=k^2\Delta_{b}.$ If the galois closure is $C_{3}$ instead of $S_{3}$, then it's even simpler since that implies that both discriminants must be squares.
More generally, for a degree $n$ extension $\mathbb{Q}(a)$ over $\mathbb{Q}$, we have that $\mathbb{Q}(a) = \mathbb{Q}(b)$ have the same Galois closures, which is at worst equal to $S_{n}$. There's $A_{n} \subseteq S_{n}$, whose fixed field is $\mathbb{Q}(\sqrt{\Delta_{a}})=\mathbb{Q}(\sqrt{\Delta_{b}})$, and so $\Delta_{a} = k^2\Delta_{b}$.
