Prove for any $\displaystyle a, b \in \mathbb{F}_{p^n}$ that if $\displaystyle x^3+ax+b$ is irreducible then $\displaystyle -4a^3-27b^2$ is a square in $\displaystyle \mathbb{F}_{p^n}$.
the splitting field of $\displaystyle p(x)=x^3 + ax +b$ is $\displaystyle \mathbb{F}_q,$ where $\displaystyle q=p^{3n}.$ let $\displaystyle \alpha, \beta, \gamma$ be the roots of $\displaystyle p(x).$ a simple calculation shows that $\displaystyle -4a^3-27b^2=[(\alpha-\beta)(\beta - \gamma)(\gamma - \alpha)]^2.$
now $\displaystyle \delta=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$ is invariant under any $\displaystyle \mathbb{F}_{p^n}$-automorphism of $\displaystyle \mathbb{F}_q.$ (why? be careful here!) thus $\displaystyle \delta \in \mathbb{F}_{p^n}. \ \Box$
Here is a more general theorem.
Let $\displaystyle F$ be a field of charachteristic not $\displaystyle 2$ and $\displaystyle f(x)\in F[x]$ be an irreducible seperable polynomial with $\displaystyle K$ the splitting field of $\displaystyle f(x)$ over $\displaystyle F$ with degree $\displaystyle n$. Let $\displaystyle \alpha_1,\alpha_2,...,\alpha_n$ be roots of $\displaystyle f(x)$ in $\displaystyle K$ (labeled in some order), define $\displaystyle \Delta = \prod_{i<j}(\alpha_i - \alpha_j)$. Any $\displaystyle \sigma \in \text{Gal}(K/F)$ can be regarded as an element of $\displaystyle S_n$. An automorphism $\displaystyle \sigma$ is even if and only if $\displaystyle \sigma (\Delta) = \Delta$ and is odd if and only if $\displaystyle \sigma (\Delta) = - \Delta$. Thus, we see from here that $\displaystyle \sigma (\Delta^2) = \Delta^2$. Therefore, $\displaystyle \Delta^2 \in K^{\text{Gal}(K/F)} = F$ because $\displaystyle K/F$ is a Galois extension.
The quantity $\displaystyle \Delta^2$ is known as the discriminant of $\displaystyle f(x)$. As NonCommAlg said it can be shown that the discriminant of $\displaystyle x^3 + ax + b$ is $\displaystyle \Delta^2 = - 4a^3 - 27b^2$. But your problem asks to show that $\displaystyle \Delta^2$ is not just in $\displaystyle F$ but it is in $\displaystyle F^2$ (the squares of $\displaystyle F$). To show this notice that if $\displaystyle \text{Gal}(K/F)\subseteq A_n$ then $\displaystyle \sigma (\Delta) = \Delta$ for each $\displaystyle \sigma \in \text{Gal}(K/F)$ and so $\displaystyle \Delta \in K^{\text{Gal}(K/F)} = F$. Therefore, $\displaystyle \Delta \in F \implies \Delta^2 \in F^2$ and we see from here that $\displaystyle -4a^3 - 27b^2 = \Delta^2 \in F^2$.
Now if $\displaystyle F = \mathbb{F}_q$ then $\displaystyle K = \mathbb{F}_{q^3}$ and the Galois extension of finite fields is cyclic which means that $\displaystyle G=\text{Gal}(K/F) \simeq \mathbb{Z}_3$. The only subgroup of $\displaystyle S_3$ which has order $\displaystyle 3$ is $\displaystyle A_3$ which means that $\displaystyle G \subseteq A_3$ and it follows by the result above that the discriminant is a square.