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Question

The value of
(1ω+ω2)(1ω2+ω4)(1ω4+ω8)(1ω8+ω16) is, where ω is the cube root of unity

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Solution

(1ω+ω2)(1ω2+ω4)(1ω4+ω8)(1ω8+ω16)
[ω3=1]
=(1ω+ω2)(1ω2+ω)(1ω+ω2)(1ω2+ω)
Using 1+ω+ω2=0,
=(ωω)(ω2ω2)(ωω)(ω2ω2)
=(2ω)(2ω2)(2ω)(2ω2)
=16ω6=16 [ω3=1]


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