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Question
Answer the following:
If ω is a complex cube root of unity, prove that (1 − ω + ω2)6 +(1 + ω − ω2)6 = 128
Solution
ω is the complex cube root of unity
∴ ω3 = 1 and 1 + ω + ω2 = 0
Also, 1 + ω2 = – ω, 1 + ω = – ω2
∴ L.H.S. = (1 – ω + ω2)6 + (1 + ω – ω2)6
= [(1 + ω2) – ω]6 + [(1 + ω) – ω2]6
= (–ω – ω)6 + (–ω2 – ω2)6
= (–2ω)6 + (–2ω2)6
= 64ω6 + 64ω12
= 64(ω3)2 + 64(ω3)4
= 64(1)2 + 64(1)4
= 128
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