`(1-w-w^2)^3+(1-w+w^2)^3`

If1,w,w^(2) be three cube roots of 1, show that: (1+w-w^(2))(1-w+w^(2))=4 and (3+w+3w^(2))^(6)=64

Show that : ( 1 + w - w^2 ) ( w + w^2 - 1 ) ( w^2 + 1 - w ) = -8

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)

(1-w^2+w^4)(1+w^2-w^4)

The value of the expression , (2-w)(2-w^2)+2.(3-w)(3-w^2)+....... ...... +(n-1).(n-w)(n-w^2) , where w is an imaginary cube root of unity is

If w is the imaginary cube root of unity evaluate |(1,w,w^2),(w,w^2,1),(w^2,1,w)|

If omega is a cube root of unity then the value of the expression 2(1+w)(1+w^(2))+3(1+w)(1+w^(2))+....+(n+1)(n+w)(n+w^(2))

Evaluate : ( 1 - 3w + w^2 ) ( 1 + w - 3w^2 )

Evaluate : ( 1 + w ) ( 1 - w^2 + w ) + w^2

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