If n is not a multiple of 3 , then n=3K+1 or n=3K+2 where K∈I Case (I) : When, n=3K+1,K∈I ωn+ω2n=ω3K+1+ω2(3K+1) =ω3k⋅ω+ω6K⋅ω2 [∵1+ω+ω2=0 and ω3=1] =(ω3)K⋅ω+(ω3)2K⋅ω2 =ω+ω2=−1 Case (II) : When, n=3K+2,K∈I ωn+ω2n=ω3K+2+ω2(3K+2)=ω3K⋅ω2+ω6K⋅ω4 =(ω3)K⋅ω2−1(ω3)2K⋅ω =ω2+ω=−1 In both cases, ωn+ω2n=−1,∀n∉ multiple of 3 .