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If ω is a complex cube root of unity, prove that (1 – ω + ω2)6 + (1 + ω – ω2)6 = 128.

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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
= R.H.S.

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