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# If ω is a complex cube root of unity, show that ωωωωa+bω+cω2c+aω+bω2 = ω2 - Mathematics and Statistics

Sum

If ω is a complex cube root of unity, show that ("a" + "b"ω + "c"ω^2)/("c" + "a"ω + "b"ω^2) = ω2

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#### 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. = ("a" + "b"ω + "c"ω^2)/("c" + "a"ω + "b"ω^2)

= ("a"ω^3 + "b"ω^4 + "c"ω^2)/("c" + "a"ω + "b"ω^2)   ...[∵ ω3 = 1, ∴ ω4 = ω]

= (ω^2("c" + "a"ω + "b"ω^2))/("c" + "a"ω + "b"ω^2)

= ω2

= R.H.S.

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