If the cube roots of unity are `1,omega,omega^2`, then the roots of the equation `(x-1)^3+8=0`, arte

If the cube roots of unity are 1,omega,omega^(2), then the roots of the equation (x-1)^(3)+8=0 are -1,1+2 omega,1+2 omega^(2) b.-1,1-2 omega,1-2 omega^(2) c.-1,-1,-1 d.none of these

If the cube roots of unity are 1,omega,omega^(2), then the roots of the equation (x-1)^(3)+8=0 are : (a)-1,1+2w,1+2w^(2)(b)-1,1-2w,1-2w^(2)(b)1,w,w^(2)

If the cube roots of unity are 1,ww^(2), then the roots of the equation (x-1)^(3)+8=0, are

If 1,omega,omega^(2) are the cube roots of unity, then the roots of the equation (x-1)^(3)+8=0 are

if 1,omega,omega^(2) root of the unity then The roots of the equation (x-1)^(3)+8=0 are

If omega ne 1 is a cube root of unity, then 1, omega, omega^(2)

If omega denotes the cube root of unity, then what is the real root of the equaton x^(3)-27=0 ?