My book says that this polynomial is irreducible over the said field. Clearly, the polynomial does not have root $r = 1$ since $(1)^3 -(1)^2 - 1 = -1 \not\equiv 0 \mod 3$.
However, for $r = -1$, we have $(-1)^3 - (-1)^2 - 1 = -3 \equiv 0 \mod 3$.
So wouldn't this mean that $x^3 - x^2 - 1$ is reducible since it has root $-1$ in $\mathbb{F}_3$?
Thanks for the help.
Yes, you are correct that it is a reducible polynomial, your book is incorrect. Since $-1$ is a root, by the Factor Theorem $(x-(-1))|(x^3-x^2-1)$ where the division is being considered in $\mathbb{Z}_3[x]$. Notably, it factors into $$(x+1)(x^2+x-1)=x^3+2x^2-1\equiv x^3-x^2-1\pmod{3}$$