hello sir. i m a bit confused while solving questions on cube roots of unity ( complex numbers ) would u plz explain me this with an example.

also would u plz help me solve this question

if omega is the cube root of unity then what is the value of (1+omega+omega^2)^7?

a. 128 omega

b. -128 omega

c. 128 omeg^2

d. -128 omega^2

Dear Nitish,

x³=1,x³-1=0,

(x-1)(x²+x+1)=0

x=1 or x²+x+1=0

x=1 or x=-(1/2)+(i√3/2) or -(1/2)-(i√3/2)

omega=-(1/2)+(i√3/2),omega^2=-(1/2)-(i√3/2)

1+omega+omega^2=sum of roots of x^3 -1=0

coefficient of x^2 in x³-1=0 is equal to zero

Therefore, 1+w+w²=0

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(1+ω+ω²)=0...........(THIS IS A PROPERTY OF THE CUBE ROOT OF UNITY)

so,(1+ω+ω²)⁷=0,

plz check ur question again,as this ans. is not among ur options.

thank u for the answer sir.

i too thought that 0 might be the answer of this but its given that w is the cibe root of unity and that confused me.
so would u plz tell me how could i solve this? 

1+w+w²=0

therefore (1+w+w²)⁷=0

there is an error in question. the actual question is (1+omega-omega^2)^7.

sol:(1+omega-omega^2)^7

=(1+omega+omega^2-2omega^2)^7

=(0-2omega^2)^7 [1+omega+omega^2=0]

=(-2omega^2)^7

=(-2)^7 x (omega^2)^7

=(-128) x (omega^14)   [omega^14=omega^2] [omega^3=1]

=(-128) x (omega^2)= -128omega^2