If ω=-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i, then ω^2=\overline{ω}, ω^3=1

ω^2+ω^4+ω^4

=\overline{ω}+ω+1

=-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i+-\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i+1

=0