Let $x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$. So $xyz=1$ and $x+y+z=-1$. So $\overline{x}+\overline{y}+\overline{z}=-1$ and hence $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-1$ since $x,y,z$ have absolute value $1$. Hence $xy+xz+yz=-1$. So $x,y,z$ are the three roots of the equation $0=t^3+t^2-t-1=(t+1)^2(t-1)$ and hence $(x,y,z)=(1,1,-1)$ up to permutations. Hence $a=b, c=-a$ up to permutations and of course all such triples satisfy the equation.