If w is a complex cube root of unity, show that (1 + w - w²)³-(1 - w + w²)³ = 0
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If w is a complex cube root of unity, show that (1 + w - w²)³-(1 - w + w²)³ = 0

[From: ] [author: ] [Date: 11-12-10] [Hit: ]
non-real).Hence,= 0.I hope this helps!......
I'm not exactly sure what I'm supposed to do. I just did difference of two cubes and was able to prove it, but I was wondering if I need work with the actual root values, i.e. (-1 + i√3)/2, etc. or if what I did was enough.

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Since w³ = 1, we have
w³ - 1 = (w - 1)(w² + w + 1) = 0
==> w² + w + 1 = 0, since w is complex (i.e., non-real).

Hence, (1 + w - w²)³ - (1 - w + w²)³
= (1 + w - (-1 - w))³ - (1 - w + (-1 - w))³
= (2 + 2w)³ - (-2w)³
= 8 [(1 + w)³ + w³]
= 8 [(1 + w) + w] [(1 + w)² - w(1 + w) + w²]
= 8 (2w + 1) [(w² + 2w + 1) - (w + w²) + w²]
= 8 (2w + 1) (w² + w + 1)
= 16 (2w + 1) * 0
= 0.

I hope this helps!
1
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