Question

If , 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 + 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