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Question
If ω is a complex cube root of unity, show that (3 + 3ω + 5ω2)6 − (2 + 6ω + 2ω2)3 = 0
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Solution
ω is a complex cube root of unity.
∴ ω3 = 1 and 1 + ω + ω2 = 0.
∴ ω + ω2 = – 1, 1 + ω = – ω2 and 1 + ω2 = – ω.
(3 + 3ω + 5ω2)6 − (2 + 6ω + 2ω2)3
= [3(1 + ω) + 5ω2]6 – [2(1 + ω2) + 6ω]3
= [3(– ω2) + 5ω2]6 - [2(– ω) + 6ω)3
= (2ω2)6 – (4ω)3
= 64ω12 – 64ω3
= 64(ω3)4 – 64ω3
= 64(1)4 – 64(1)
= 0
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