Question

If `alpha` be the real cube root of and `beta`, `gamma` be the complex cube roots of `m`, a real positive number, then for any `x`, `y`, `z` show that `(xbeta+ygamma+zalpha)/(xgamma+yalpha+zbeta)=omega^2`, where `omega` is a complex cube root of unity.

Answer 1

`omega^2+ omega + 1 = 0`
`omega^3 = 1`
`root(3)(m) *1 = alpha`
`root(3)(m) *omega= beta`
`root(3)(m) *omega^2 = gamma `
`=(root(3)(m) { xomega + yomega^2 + z})/(root(3)(m) {x omega^2 + y + z omega})`
`=(x omega + omega^2y+z)/(xomega^2 + y +z omega)*(omega^2/omega^2)`
`= ((x omega + omega^2y+z)*omega^2)/(xomega^4 + yomega^2 +z omega^3)`
`= (x omega + omega^2y + z)/(x omega + y omega^2 + z) * omega^2`
`=1 * omega^2 = omega^2`