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Question
If ω is a complex cube root of unity, show that (1 + ω − ω2)6 = 64
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. = (1 + ω − ω2)6
= [(1 + ω) − ω2]6
= (−ω2 − ω2)6
= (−2ω2)6
= 64.ω12
= 64(ω3)4
= 64(1)4
= 64
= R.H.S.
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