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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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Question

If , where α and β are the complex cube-roots of unity, show that xyz = a3 + b3.

Sum
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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 + isqrt3)/2`  and  β = `(-1 - isqrt3)/2`

∴ αβ = `((-1 + isqrt3)/2)((-1 - isqrt3)/2)`

= `((-1)^2 - (isqrt3)^2)/4`

= `(1-(-1)(3))/4`             ...[∵ i2 = -1]

= `(1 + 3)/4`

= `4/4`

∴ αβ = 1

Also, α + β = `(-1 + isqrt3)/2 + (-1 - isqrt3)/2`

= `(-1 + isqrt3 -1 - isqrt3)/2`

= `-2/2`

α + β = −1

∴ 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

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Chapter 1: Complex Numbers - Exercise 1.4 [Page 20]

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