Given that ω=cos2π3+isin2π3
Therefore, 1+ω+ω2=0 & ω3=1
∣∣
∣
∣∣z+1ωω2ωz+ω21ω21z+ω∣∣
∣
∣∣=0
R1⟶R1+R2+R3
∣∣
∣
∣∣z+1+ω+ω2z+1+ω+ω2z+1+ω+ω2ωz+ω21ω21z+ω∣∣
∣
∣∣=0
⇒z∣∣
∣
∣∣111ωz+ω21ω21z+ω∣∣
∣
∣∣=0
⇒z[(z+w2)(z+w)−1−ω(z+ω)+ω2−ω2(z+ω2)]=0
⇒z[z2+z(ω+ω2)−z(ω+ω2)]=0
⇒z3=0
Ans: 1