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Question

If 1, ω, ω2 be the three cube roots of unity, then
(1+ω)2n1n=1(1+ω2n)=

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Solution

Given (1+ω)2n1n=1(1+ω2n)
Expanding, we get (1+ω)(1+ω2)(1+ω4)(1+ω8)(1+ω22n1) 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)2n terms
=1 [ω3=1]

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