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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
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Solution
ω is the 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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