Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 4 / 6E / d / Solution

Factor the equation asFor the nontrivial roots, the quadratic formula over residues givesThe supplied square root is $14$, and $2^{-1}\equiv 100(\mathrm{mod}199)$. ThereforeIndeed $107^2-107+1=11343=57\cdot 199$, so $107^3+1$ is divisible by $199$, and $107\not\equiv -1$. The other nontrivial root is $(1-14)100\equiv 93$; it satisfies $93^2-93+1=43\cdot 199$. 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 $(1\pm \sqrt{-3})/2$.