When does $x^3 - x^2 - 2 x + 1$ split mod $p$? Conjecture: the primes for which the polynomial $x^3 - x^2 - 2 x + 1$ splits mod $p$ are the primes $\equiv 1$ or $6$ mod $7$ (OEIS sequence A045472).  Is this correct?
If $\alpha$ is a root of $x^3-x^2-2x+1$ (in any field) then the latter splits as $$(x-\alpha)(x-\alpha^2+\alpha+1)(x+\alpha^2-2).$$ Also if $\alpha \neq -2$ and $\beta$ is a root of $$x^2+\alpha x + 1$$ then $\beta$ is a primitive seventh root of unity. Combining all this for the field $\mathbb{F}_p$ where $p\neq 7$: If $x^3-x^2-2x+1$ has a root in $\mathbb{F}_p$ then $\mathbb{F}_{p^2}$ contains a primitive seventh root of unity and so $p^2 \equiv 1 \pmod 7$. The other way around: If $p^2\equiv 1 \pmod 7$ then $\mathbb{F}_{p^2}$ contains a primitive seventh root of unity $\beta$ and $\alpha = -\beta-\beta^{-1}$ is a root of $x^3-x^2-2x+1$. Moreover $\beta^p = \beta$ (if $p\equiv 1 \bmod 7$) or $\beta^p = \beta^{-1}$ (if $p \equiv -1 \bmod 7)$. Either way $\alpha^p = \alpha$ so $\alpha \in \mathbb{F}_p$. For $p=7$ $$x^3-x^2-2x+1\equiv (x+2)^3 \pmod 7$$ as pointed out by @lhf in a comment.
Yes, this is because the zeros of the polynomial is $-(\zeta^j+\zeta^{-j})$
for $\zeta=\exp(2\pi i/7)$ and $j\in\{1,2,3\}$. So it splits iff $\zeta+\zeta^{-1}\in\Bbb F_p$ for $\zeta$ now a primitive seventh root of unity
in an extension of $\Bbb F_p$. This is the case iff $(\zeta+\zeta^{-1})^p
=\zeta+\zeta^{-1}$ in characteristic $p$, and that is the case iff
$p\equiv\pm1\pmod7$.