If ω≠1 is a cube root of unity and ∣∣ ∣ ∣∣x+ω2ω1ωω21+x1x+ωω2∣∣ ∣ ∣∣=0, then value of x is
0
Let Δ=∣∣
∣
∣∣x+ω2ω1ωω21+x1x+ωω2∣∣
∣
∣∣
Applying C1→ C1+C2+C3 we get
Δ=∣∣
∣
∣∣x+ω2+ω+1ω1ω+ω2+1+xω21+x1+x+ω+ω2x+ωω2∣∣
∣
∣∣=∣∣
∣
∣∣xω1xω21+xxx+ωω2∣∣
∣
∣∣ (∵1+ω+ω2=0)
Δ is clearly equal to 0 for x = 0.