We define the polynomial $f(T) := T^6 - 3T^2 + 1 \in \mathbb Q[T].$ Let $\alpha \in \mathbb C$ be a root of $f.$
I showed that $f \mod 3$ can be factorized into three irreducible factors of degree $2:$ $$f(T) \mod 3 = T^6 + 1 = (T^2)^3 + 1^3 = (T^2 + 1)^3$$
Now I am supposed to see that the degree of the minimal polynomial of $\alpha$ over $\mathbb Q$ has to be even, but I fear I don't get it at the moment. (I know that $f$ is in fact irreducible and therefore the minimal polynomial of degree $6$, but I don't want to use that reasoning.)
In general, you want to show that if a polynomial has an odd-degree factor over $\Bbb Q$, then it must have an odd-degree factor modulo any prime as well (the contrapositive will then give you what you want). This can be verified simply by looking at the factors of $f$ over $\Bbb Q$ and how each one factors modulo $3$.
Given a polynomial $g$ of degree $d$ over a field $F$, its factorization into irreducibles of degrees $d_j$ gives a partition of $d$. The above claim follows from an even more general statement: the partition yielded by the factorization of $g$ modulo any prime is a refinement of the partition yielded by the factorization of $g$ over $\Bbb Q$. This can be verified simply by looking at each irreducible factor of $g$ over $\Bbb Q$ and how it factors modulo the prime.