If 1,ω,ω2 be the three cube roots of unity, then (1+ω)2n−1∏n=1(1+ω2n)=
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Solution
Given (1+ω)2n−1∏n=1(1+ω2n)
Expanding, we get (1+ω)(1+ω2)(1+ω4)(1+ω8)⋯(1+ω22n−1) upto 2n terms (1+ω)(1+ω2)(1+ω)(1+ω2)⋯⋯(1+ω2) upto 2n terms
Using property 1+ω+ω2=0 =(−ω2)(−ω)(−ω2)(−ω)⋯2nterms =1[∵ω3=1]