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Question
If ω is a complex cube root of unity, show that (a − b) (a − bω) (a − bω2) = a3 − b3
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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. = (a − b)(a − bω)(a − bω2)
= (a − b) (a2 − abω2 − abω + b2ω3)
= (a − b) [a2 − ab(ω2 + ω) + b2(1)]
= (a − b) [a2 − ab(−1) + b2]
= (a − b) (a2 + ab + b2)
= a3 − b3
= R.H.S.
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