Pairwise distinctness makes the cyclic difference product nonzero, so subtracting the cyclic equations and cancelling it gives `(a+b)(b+c)(c+a)=1`. With `s=a+b+c`, the original equations give `a^2+b^2+c^2=s+18`; eliminating the other symmetric terms yields `(2s-3)(s^2+s-4)=0`. On the branch `s=3/2`, the pair sum is `-69/8` and the product is `-223/16`. The original system also requires `(a^2-6)(b^2-6)(c^2-6)=abc`, but these invariants give the two sides as `-4343/256` and `-223/16`, a nonzero residual `-775/256`. Thus that branch is extraneous. The remaining values are `(-1-sqrt(17))/2` and `(-1+sqrt(17))/2`; in particular, the proposed `-3/2` fails the necessary polynomial.
