`omega^3` = 1
`|(x+1,omega,omega^2),(omega,x+omega^2,1),(omega^2,1,x+omega)| = 0`
then value of x is?
(a)`0`
(b)`omega`
(c)`omega^2`
(d)None

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

If omega is a cube root of unity , then |(x+1 , omega , omega^2),(omega , x+omega^2, 1),(omega^2 , 1, x+omega)| =

|[omega+omega^(2),1,omega],[omega^(2)+1,omega^(2),1],[1+omega,omega,omega^(2)]|

If omega!=1 is a cube root of unity and Delta=|(x+omega^(2),omega,1),(omega,omega^(2),1+x),(1,x+omega,omega^(2))|=0 then value of x is

{[(1,omega,omega^(2)),(omega,omega^(2),1),(omega^(2),1,omega)] + [(omega,omega^(2),1),(omega^(2),1,omega),(omega,omega^(2),1)]} [(1),(omega),(omega^(2))]

If omega is a cube root of unity, then for polynomila is |(x + 1,omega,omega^(2)),(omega,x + omega^(2),1),(omega^(2),1,x + omega)|