Home
Class 14
MATHS
[" If "omega" is the cube root of unity ...

[" If "omega" is the cube root of unity then the polynomian "],[|[x+1,omega,omega^(2)],[omega,x+omega^(2),1],[omega^(2),1,x+omega]|]

Promotional Banner

Similar Questions

Explore conceptually related problems

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)|

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

If omega is a complex cube root of unity then a root of the equation |[x+1,omega,omega^2],[omega,x+omega^2,1],[omega^2,1,x+omega]|=0 is

If omega is a cube root of unity then the value of |(1,omega,omega^(2)),(omega,omega^(2),1),(omega^(2),1,omega)| is -

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

If omega is a complex cube root of unity, then a root of the equation |(x +1,omega,omega^(2)),(omega,x + omega^(2),1),(omega^(2),1,x + omega)| = 0 , is

If omega is a complex cube root of unity, then a root of the equation |(x +1,omega,omega^(2)),(omega,x + omega^(2),1),(omega^(2),1,x + omega)| = 0 , is

If omega is cube root of unity, then a root of the equation |(x + 1,omega, omega^2),(omega, x + omega , 1),(omega^2 , 1, x + omega)| = 0 is