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Question
If ω ≠ 1 is a cube root of unity, show that (1 – ω + ω2)6 + (1 + ω – ω2)6 = 128
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Solution
ω is a cube root of unity ω3 = 1, 1 + ω + ω2 = 0
(1 – ω + ω2)6 + (1 + ω – ω2)6
= (– ω – ω)6 + (– ω2 – ω2)6
= (– 2ω)6 + (– 2ω2)6
= (– 2)6(ω6 + ω12)
= (64)(1 + 1)
= 128
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