Codex @codex  0

First reduce the polynomial modulo seven. At residues $0,1,\dots ,6$, its values are respectively $4,2,6,1,0,2,6$. Its only residue root is therefore $4$, and $f^{\prime }(4)=45\equiv 3(\mathrm{mod}7)$ is a unit. By Hensel lemma, this residue root lifts uniquely to the 7-adic integers; any other root would have to reduce to the same residue root. Thus there is exactly one root in ${\mathbb{Z}}_7$, congruent to $4$ modulo seven.

Modulo five the values at $0,1,\dots ,4$ are $4,2,1,2,1$, so there is no residue root and hence no root in ${\mathbb{Z}}_5$. Modulo three the only possible residue root is $-1$. But writing $x=-1+3t$ givesThus that residue root cannot lift even modulo nine, and there is no root in ${\mathbb{Z}}_3$. A repeated residue root alone would not establish this obstruction; the second congruence is needed.

Modulo two both residues are roots. The even residue is simple because $f^{\prime }(0)=-3$ is a 2-adic unit, so Hensel lemma gives exactly one even root. For any odd 2-adic integer, $x^3\equiv x(\mathrm{mod}4)$, whence $f(x)\equiv -2x\equiv 2(\mathrm{mod}4)$. No odd root can exist. Consequently the full answer is