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Question

If 1,ω and ω2 are the cube roots of unity, prove that a+bω+cω2c+aω+bω2=ω2

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Solution

Given, a+bω+cω2c+aω+bω2

=ω2(a+bω+cω2)ω2(c+aω+bω2)

=ω2(a+bω+cω2)cω2+aω3+bω4

=ω2(a+bω+cω2)(cω2+a+bω) (ω3=1)

=ω2

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