If 1, ω, ω2 are the cube roots of unity, then Δ= |1 ωn ω2 n ωn ω2 n 1 ω2 n 1 ωn| is equal to-

Question

If 1, ω, ω2 are the cube roots of unity, then Δ= |1 ωn ω2 n ωn ω2 n 1 ω2 n 1 ωn| is equal to- (A) 0 (B) 1 (C) ω (D) ω2

Tardigrade Admin · Accepted Answer

Given Δ=|1 ωn ω2 n ωn ω2 n 1 ω2 n 1 ωn|=1(ω3 n-1)-ωn(ω2 n-ω2 n)+ω2 n(ωn-ω4 n) =1(1-1)-0+ω2 n(ωn-ωn)=0

Q. If $1, ω, ω^{2}$ are the cube roots of unity, then $Δ = ∣ ∣ 1 ω^{n} ω^{2 n} ω^{n} ω^{2 n} 1 ω^{2 n} 1 ω^{n} ∣ ∣$ is equal to-