polynomial in finite field

Here is a more general theorem.

Let $F$ be a field of charachteristic not $2$ and $f(x)\in F[x]$ be an irreducible seperable polynomial with $K$ the splitting field of $f(x)$ over $F$ with degree $n$ . Let $\alpha_1,\alpha_2,...,\alpha_n$ be roots of $f(x)$ in $K$ (labeled in some order), define $\Delta = \prod_{i<j}(\alpha_i - \alpha_j)$ . Any $\sigma \in \text{Gal}(K/F)$ can be regarded as an element of $S_n$ . An automorphism $\sigma$ is even if and only if $\sigma (\Delta) = \Delta$ and is odd if and only if $\sigma (\Delta) = - \Delta$ . Thus, we see from here that $\sigma (\Delta^2) = \Delta^2$ . Therefore, $\Delta^2 \in K^{\text{Gal}(K/F)} = F$ because $K/F$ is a Galois extension.

The quantity $\Delta^2$ is known as the discriminant of $f(x)$ . As NonCommAlg said it can be shown that the discriminant of $x^3 + ax + b$ is $\Delta^2 = - 4a^3 - 27b^2$ . But your problem asks to show that $\Delta^2$ is not just in $F$ but it is in $F^2$ (the squares of $F$ ). To show this notice that if $\text{Gal}(K/F)\subseteq A_n$ then $\sigma (\Delta) = \Delta$ for each $\sigma \in \text{Gal}(K/F)$ and so $\Delta \in K^{\text{Gal}(K/F)} = F$ . Therefore, $\Delta \in F \implies \Delta^2 \in F^2$ and we see from here that $-4a^3 - 27b^2 = \Delta^2 \in F^2$ .

Now if $F = \mathbb{F}_q$ then $K = \mathbb{F}_{q^3}$ and the Galois extension of finite fields is cyclic which means that $G=\text{Gal}(K/F) \simeq \mathbb{Z}_3$ . The only subgroup of $S_3$ which has order $3$ is $A_3$ which means that $G \subseteq A_3$ and it follows by the result above that the discriminant is a square.