 # if w(omaga) is an imaginary cube root of unity, then (1+w - w2)7 is equal to?.......................please write full answer to more know to me........................note:-w2(it square on omega) and (1+ w -w2)7 [it is square on w(omega) and whole root 7 on (1+w-w2)7]

10 years ago

Hi Rahul,

As w is a cube root of unity,

We have

1+w+w2 = 0.

So 1+w = -w2

Now 1+w-w2 = -w2-w2 = -2w2.

So (1+w-w2)7 = (-2w2)7 = (-2)7w14.

We also know that w3 = 1.

So (w3)2 also = 1

Similarly w12=1

So finally we have w14 = w2.

Hope that helps.

All the best.

Regards,

one year ago

As we know that

If 1,w,w2 is the cube roots of unity then

(1+w+w2)=0 and w3=1

then 1+w2 = -w

1+w = -w2

According to question we have to find (1 + w2 - w) (1 - w2 + w)

So first we have to find (1 + w2 - w) and (1 - w2 + w)

then,

putting 1+w2 = -w in (1 + w2 - w)

(1 + w2 - w) = (-w-w)

(1 + w2 - w) = (-2w)…………………………...(Eq(1))

Again,

putting 1+w = -w2 in (1 - w2 + w)

(1 - w2 + w) = (-w2-w2)

(1 - w2 + w) = (-2w2)……………………….(Eq(2))

so,

(1 + w2 - w) (1 - w2 + w) = (Eq(1)) x (Eq(2)) = (-2w) x (-2w2) = ( 4w3)

As we know that w3=1

So,

(1 + w2 - w) (1 - w2 + w) = ( 4w3) = (4 x1)= 4

Hence,

(1 + w2 - w) (1 - w2 + w) = 4

Thank you :)