`1+omega+omega^2 = 0`

If 1, omega, omega^2 be three roots of 1, show that: (1+omega)^3-(1+omega^2)^3=0

If 1, omega, omega^2 be three roots of 1, show that: (1-omega+omega^2)^2+(1+omega-omega^2)^2=-4

det [[1, omega, omega^(2) omega, omega^(2), 1omega^(2), 1, omega]]

det [[1, omega, omega^(2) omega, omega^(2), 1omega^(2), 1, omega]] =

If 1, omega, omega^2 ,…..,omega^(n-1) are the nth roots of unity, then (2- omega)(2-omega^2)….(2-omega^(n-1) ) equals

If omega is a cube root of unity, prove that (1+omega-omega^2)^3-(1-omega+omega^2)^3=0

|[1,omega,omega^2] , [omega, omega^2,1] , [omega^2,1,omega]|=0

If 1 , omega , omega^(2) are the cube roots of unity then the values ((a+b omega+c omega^(2))/(c+a omega+b omega^(2))) + ((a+b omega+c omega^(2))/(b+c omega+a omega^(2))) = (A) 1 (B) 0 (C) -1 (D) None of these