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Question
If , 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 + 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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