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Question
If x = a + b, y = αa + βb and z = aβ + bα, where α and β are the complex cube roots of unity, show that xyz = a3 + b3.
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Solution
x = a + b, y = αa + βb and z = aβ + bα
α and β are the complex cube roots of unity.
∴ α = `(-1 + "i"sqrt(3))/2 and beta = (-1 - "i"sqrt(3))/2`
∴ αβ = `((-1 + "i"sqrt(3))/2)((-1 - "i"sqrt(3))/2)`
= `((-1)^2 - ("i"sqrt(3))^2)/4`
= `(1 - (-1)(3))/4` ...[∵ i2 = – 1]
= `(1 + 3)/4`
∴ αβ = 1
Also, α + β = `(-1 + "i"sqrt(3))/2 + (-1 - "i"sqrt(3))/2`
= `(-1 + "i"sqrt(3) - 1 - "i"sqrt(3))/2`
= `(-2)/2`
∴ α + β = -1
L.H.S. = xyz = (a + b)(αa + βb)(aβ + bα)
= (a + b)(αβa2 + α2ab + β2ab + αβb2)
= (a + b)[1. (a2) + (α2 + β2)ab + 1. (b2)]
= (a + b) {a2 + [(α + β)2 – 2αβ ]ab + b2}
= (a + b) {a2 + [(– 1)2 – 2(1)]ab + b2}
= (a + b) [a2 + (1 – 2)ab + b2]
= (a + b)(a2 – ab + b2)
= a3 +b3
= R.H.S.
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