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Question
If ω is a complex cube root of unity, show that (a + b)2 + (aω + bω2)2 + (aω2 + bω)2 = 6ab
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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.
(a + b)2 + (aω + bω2)2 + (aω2 + bω)2
= (a2 + 2ab + b2) + (a2ω2 + 2abω3 + b2ω4) + (a2ω4 + 2abω3 + b2ω2)
= a2 + 2ab + b2 + a2ω2 + 2ab + b2ω + a2ω + 2ab + b2ω2 ...[∵ ω4 = ω3 × ω = ω]
= 6ab + (a2 + a2ω2 + a2ω) + (b2 + b2ω + b2ω)
= 6ab +a2(1 + ω + ω2) + b2(1 + ω + ω2)
= 6ab + a2(0) + b2(0)
= 6ab
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