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Question

Given that 1,ω,ω2 are cube roots of unity, show that (1ω+ω2)5+(1+ωω2)5=32.

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Solution

Given,

(1w+w2)5+(1+ww2)5

1+w+w2=0,w3=1

=(ww)5+(w2w2)5

=(2w)5+(2w2)5

=(2)5(w+w2)5

=(32)(1)5

=(32)(1)

=32

Hence proved.

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