Minimal polynomial?

Let $$K$$ a field, $$p,n\in \mathbb{N}$$, $$B\in \mathcal{M}_p({K})$$ and let denote $$S_B=\{X\in \mathcal{M}_p(K) \ \mid \ X^n=B\}$$.

If $$X\in S_B$$, I have to prove that $$\mu_X$$ (the minimal polynomial of $$X$$) divides $$\mu_B(\xi^n)$$.

If $$X\in S_B$$ then we can deduce that $$X^n-B=0$$ but after I do not know how to link this with $$\mu_X$$ or $$\mu_B$$...

Hint: It suffices to note that if $$X \in S_B$$, then $$\mu_B(X^n) = 0$$
• Does it come from the fact that $\mu_B(B)=0$ by definition ? – Maman Jan 11 at 19:15
• Then I think we have : $\mu_{X^n}(X^N)=0$ ? What is the link with $\xi^n$ ? – Maman Jan 12 at 21:27
• If $p(A)=0$, then the minimal polynomial of $A$ divides $p$. – Omnomnomnom Jan 13 at 1:49
• @Maman $\xi$ is used to emphasize the polynomial itself. That is, $p(\xi) = \mu_B(\xi^n)$ is a polynomial satisfying $p(X) = 0$ – Omnomnomnom Jan 13 at 16:54