This exercise originates from Pinter's Abstract Algebra, Chapter 31, B1.
Show that, in any extension of $\mathbb{Z}_3$ which contains a root $u$ of \begin{align*} a(x) &= x^3 + 2x + 1 \in \mathbb{Z}_3[x] \end{align*} it happens that $u + 1$ and $u + 2$ are the remaining two roots of $a(x)$. Use this fact to find the root field of $x^3 + 2x + 1$ over $\mathbb{Z}_3$. List the elements of the root field.
Given $a(u)=0, a(u+1)=(u+1)^3 + 2(u+1) + 1 = (u^3+1)+(2u+2)+1 = a(u) = 0$.
Similarly, $a(u+2)=(u+2)^3 + 2(u+2) + 1 = (u^3+2) + (2u+1) + 1 = a(u) = 0$.
This shows $u,u+1,u+2\in\mathbb{Z}_3(u)\implies\mathbb{Z}_3(u)$ is the root field of $x^3 + 2x + 1$ over $\mathbb{Z}_3$.
Viewed as a vector space with basis $\{1, u, u^2\}$ over $\mathbb{Z}_3, \mathbb{Z}_3(u)$ has a total of $3^3=27$ elements: $\qquad\qquad\qquad\qquad\begin{array}{r r r} \\ 0, & 1, & 2, \\ u, & u+1, & u+2, \\ 2u, & 2u+1, & 2u+2, \\ u^2, & u^2+1, & u^2+2, \\ u^2+u, & u^2+u+1, & u^2+u+2, \\ u^2+2u, & u^2+2u+1, & u^2+2u+2, \\ 2u^2, & 2u^2+1, & 2u^2+2, \\ 2u^2+u, & 2u^2+u+1, & 2u^2+u+2, \\ 2u^2+2u, & 2u^2+2u+1, & 2u^2+2u+2 \\ \end{array}$
Correct?
Edit: I just noticed that you haven't shown that the polynomial is irreducible. This is very easy to do - just check that it has no roots, and use the fact that a cubic polynomial with no roots is irreducible.
Yep, looks good! Just a small note for extra context:
You've shown that the root field is a field with $27$ elements. In fact, for any prime $p$ and integer $n \geq 1$, there is exactly one field of size $p^n$, commonly denoted $\mathbb{F}_{p^n}$ or $\text{GF}(p^n)$ for "Galois Field", and this is a field extension of $\mathbb{Z}_p$ of degree $n$. You can actually prove this without too much machinery - I'd recommend looking it up if you're interested. Thus, when the question asks you to find "the" root field, I think it would probably be natural to say that the answer is $\mathbb{F}_{3^3}$, since this actually tells us exactly what the field is up to isomorphism. An immediate corollary is that actually the root field is the same for any irreducible cubic.
Apologies if you already knew this - just thought I'd throw it out there in case you didn't ^_^