$X^3-2$ splits completely in an extension of $\mathbb F_7$ My question concerns the following problem:

Let $K=\mathbb F_7[T]/(T^3-2)$. Show that $X^3-2$ splits into linear factors in $K[X]$.

Write $K\simeq \mathbb F_7[\alpha]$ for a root $\alpha\in \overline{\mathbb F_7}$ of $X^3-2$. I found out through experimentation that $2\alpha$, $4\alpha$ are also roots of this polynomial (since $2^3\equiv 4^3 \equiv 1 \ \mathrm{mod}\ (7)$). Is there a more conceptual way to see this?

$X^3-2$ is irreducible mod $7$. This implies that $K$ has $7^3$ elements.
Since the group $K^\times$ is cyclic of order $7^3-1= 2 \cdot 3^2 \cdot 19$, there is an element $\omega \in K^\times$ of order $3$.
Then $X^3-2=(X-\alpha)(X-\omega\alpha)(X-\omega^2\alpha)$.
As you have discovered, you can take $\omega=2$.
