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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]
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.
And hence answer is -27w2
Hope that helps.
All the best.
Regards,
Ashwin (IIT Madras).
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 :)
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