Number of irreducible cubics over a field of $n$ elements.

Question

This question originates from Pinter's Abstract Algebra, Chapter 25 Exercise C4.

How many irreducible cubics are there over a field of $n$ elements?

1. Total number of monic cubics $x^3 + ax + bx + c$: $n^3$.

2. For monic reducible cubics in the form of $(x+a)(x+b)(x+c)$, either all, only two, or none of $a, b, c$ are the same. Therefore the total number of such cases is $\displaystyle {n\choose 1} + 2{n\choose 2} + {n\choose 3}$.

3. Per this discussion, the number of monic irreducible quadratics is $n(n-1)/2$. This suggests for monic reducible cubics in the form of $(x^2+ax+b)(x+c)$ where $x^2+ax+b$ is irreducible, there are $n^2(n-1)/2$ cases.

Hence, the total number of

Monic reducible cubics:

$\displaystyle\quad R = {n\choose 1} + 2{n\choose 2} + {n\choose 3} + \frac{n^2(n-1)}{2}$

Monic irreducible cubics:

$\displaystyle\quad n^3 - R$

Irreducible cubics:

$\displaystyle\quad(n-1)\left(n^3 - R\right)$.

Correct?