Is there any argument with the minimal polynomial for this question? Let $f\in \mathbb{Q}[X]$ be a polynomial which has the root $\sqrt[3]2$. Prove that the polynomial $f$ is divisible by $X^3-2$.
I have just started working on polynomials, so I just said that from the Euclidean division we know that $\exists q, r \in \mathbb{Q}[X]$ such that $f=(X^3-2)q+r$, where $r=aX^2+bX+c$, $a, b, c\in \mathbb{Q}$.
By plugging in $\sqrt[3]2$, we get that $a\sqrt[3]4 +b\sqrt[3] 2+c=0$, so $a=b=c=0$, which means that $r=0$ and we reach our conclusion.
Now, I know that $\sqrt[3]2$'s minimal polynomial over $\mathbb{Q}[X]$ is precisely $X^3-2$, so I was wondering if some more general result regarding the minimal polynomial holds.

The polynomial $X^3-2$ is irreducible over $\mathbb Q$, since otherwise it would have a linear factor, which would give a rational root. Alternatively you can also apply Eisenstein criterion with the prime $2$.
Since $f$ and $X^3-2$ have a common root, their $\gcd$ has positive degree. Since the $\gcd$ is in $\mathbb{Q}[X]$, as computed by the Euclidean algorithm, it must be a rational polynomial factor of $X^3-2$ of positive degree. Since this is irreducible, then that factor must be all of $X^3-2$. Hence $X^3-2$ divides $f$.
