Question

If W is a complex cube root of unity, show that (1+W–W²)^6=64​

Answer 1

Answer:

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Step-by-step explanation:

If w is a cube roots of unity then we have (i) w^3 = 1 and (ii) 1+w+w^2 = 0 . Therefore 1 + w - w^2=-w^2 -w^2 = -2w^2 which in turn gives (1+w-w^2)^6 = (-2w^2)^6 = 64w^(12) = 64 .

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