Following is the whole question:
If $a$ and $b$ are elements of a finite field $k$, then $f(x)$ must have a root in $k$, where $f(x)=(x^{3}+ax+b)(x^{2}+4a^{3}+27b^{2})$.
I have finished most of the proof, but I was stuck in the near end.
First, assume $char(k)=p>0 \neq 2,3$. Suppose $f(x)$ has no root in $k$.
Then, the discriminant of $g(x)=(x^{3}+ax+b)$ is not a square in $k$. This is true if and only if the Galois group of $g$ over k is $S_{3}$. However, this cannot be true since any Galois extension over a finite field is cyclic.
However, I am now stuck in the case where $char(k)=2$ or $3$. I am learning algebra following Lang's book, so in the argument of the discriminant of the cubic extensions and the Galois group, he assumes that $char(k)$ is not $2$ or $3$. Therefore, I don't know how to argue in the next steps.
Any thoughts and explanations are really appreciated!!
EDIT 1
Let me modify my proof a little bit, since my original proof is not totally correct. Note that even after the modification, I still have the original question, since in this proof I used $char(k)\neq 2$ or $3$
$f(x)$ has must have a root in $k$ implies, either $g(x)=x^{3}+ax+b$ has a root or $h(x)=x^{2}+4a^{3}+27b^{2}$ or they have a common root.
My proof above eliminate the chance that they have a common root.
So now, suppose $g(x)$ is irreducible over $k=\mathbb{F_{p^{n}}}$.
As $g(x)$ is irreducible over $k$ of degree 3, the Galois group of $g(x)$ over $k$ has order 3 or 6 since the Galois group if a subgroup of $S_{3}$. However, since every Galois extension over a finite field is cyclic, it must be of order 3, which implies that the Galois group is $A_{3}$.
This implies that the discriminant of $g(x)$ is a square in $k$, but notice that the discriminant of $g(x)$ is $-4a^{3}-27b^{2}$, which is the solution of $h(x)=0$. Since it is a square, $h(x)=0$ which implies $x^{2}=-4a^{3}-27b^{2}$ has a solution.
Thus, there must be always solution for $f(x)$,
However, I still do not know how to deal with the case where $char(k)=2$ or $3$.
EDIT 2
Now, with my quasi's answer and my edited proof, this whole proof is complete!
If $\text{char}(k)=2$, then $x=b$ is a root of $x^{2}+4a^{3}+27b^{2}$.
Next, suppose $\text{char}(k)=3$.
If the function $g:k\to k$ given by $g(x)=x^3+ax+b$ is injective, then since $k$ is finite, $g$ would also be surjective, hence $g(x)$ would equal $0$, for some $x\in k$.
Suppose then that $g$ is not injective.
Thus, for some $u,v\in k$, with $u\ne v$, we have $g(u)=g(v)$. \begin{align*} \text{Then}\;\;&g(u)=g(v)\\[4pt] \implies\;&u^3+au+b=v^3+av+b\\[4pt] \implies\;&u^3-v^3=-a(u-v)\\[4pt] \implies\;&u^2+uv+v^2 = -a\\[4pt] \implies\;&u^2-2uv+v^2 = -a\\[4pt] \implies\;&(u-v)^2= -a\\[4pt] \end{align*} Let $t=u-v$, and let $x=t^3$. \begin{align*} \text{Then}\;\;x^{2}+4a^{3}+27b^{2}&=x^{2}+a^{3}\\[4pt] &=t^6+a^3\\[4pt] &=(t^2)^3+a^3\\[4pt] &=(-a)^3+a^3\\[4pt] &=0 \end{align*}