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Question
If ω is a complex cube root of unity, show that (2 + ω + ω2)3 - (1 - 3ω + ω2)3 = 65
Solution
ω is a complex cube root of unity.
∴ ω3 = 1 and 1 + ω + ω2 = 0
Also, 1 + ω2 = - ω, 1 + ω = - ω2 and ω + ω2 = - 1
L.H.S. = (2 + ω + ω2)3 - (1 - 3ω + ω2)3
= [(2 + (ω + ω2)]3 - [(- 3ω + (1 + ω2)]3
= (2 - 1)3 - (- 3ω - ω)3
= 13 - (- 4ω)3
= 1 + 64ω3
= 1 + 64(1) = 65
= R.H.S.
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