Roots of a cubic mod prime For which primes $p$ is there a root to the equation $x^3+x^2-2x-1$ mod $p$? I have no idea where to start, any help is appreciated! Thank you

This is a bit of a trick question, for the following reason. First, if we approach this question "honestly", and ask about generic cubics, there is not much one can say at an elementary level, in part (indirectly) because the Galois group over $\mathbb Q$ is probably not abelian (so, secretly, "classfield theory", the well-developed study of questions of this sort for abelian extensions would not apply). 
However, since the question is asked at all, one might suspect that the Galois group over $\mathbb Q$ is abelian. Both because one is disinclined to compute a discriminant of a cubic, and because one suspects that the polynomial is special, anyway, my reaction is to wonder whether it's the simplest cubic I know with abelian Galois group over $\mathbb Q$, namely, that for the cubic subfield of the field of seventh roots of unity (with cyclic Galois group of order $6$, so admitting a unique cubic subfield).
Indeed, a standard trick going back at least 240 years: from $x^6+x^5+\ldots+x+1=0$, dividing through by $x^3$, gives $x^3+x^2+x+1+x^{-1}+x^{-2}+x^{-3}=0$. Letting $y=x+x^{-1}$, we find $y^3+y^2-2y-1=0$. [Edit: terrible typo: the $y^2$ term was earlier written just as $y$. Sorry!]
Thus, that cubic factoring means there is a linear factor, so a seventh root of unity is at most quadratic over $\mathbb F_p$. That is, either there is a seventh root of $1$ in $\mathbb F_p$ already, which is $7|(p-1)$, or in the quadratic extension, so $7|(p^2-1)$. The latter condition subsumes the former, so the condition is $7|(p^2-1)$, which is $p=\pm 1\mod 7$, since $7$ is prime.
Edit-edit: as in commments by Will Jagy, the cubic $x^3+x^2-4x+1$ apparently is a cubic with roots in the unique cubic subfield of 13th cyclotomic field. :)
Edit-edit-edit: indeed, as Gerry M notes, the 9th roots of unity have an arguably even simpler cubic subfield. And/but we'd recognize that cubic, indeed. Maybe future generations will all recognize the cubic subfields of 7th and 13th roots. :)
