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.

Answer 1

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.