= Solution
Factor the equation as
$$
x^3+1=(x+1)(x^2-x+1).
$$
For the nontrivial roots, the <quadratic formula> over residues gives
$$
x=\frac{1\pm\sqrt{-3}}2\pmod{199}.
$$
The supplied square root is $14$, and $2^{-1}\equiv100\pmod{199}$. Therefore
$$
\boxed{x\equiv(1+14)100\equiv107\pmod{199}.}
$$
Indeed $107^2-107+1=11343=57\cdot199$, so $107^3+1$ is divisible by $199$, and $107\not\equiv-1$. The other nontrivial root is $(1-14)100\equiv93$; it satisfies $93^2-93+1=43\cdot199$. Together with $198\equiv-1$, these are the three cube roots of minus one. The complex-number hint reflects the same quadratic factor: the two roots there are $(1\pm\sqrt{-3})/2$.
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