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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]

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]

Grade:12th Pass

2 Answers

Ashwin Muralidharan IIT Madras
290 Points
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.

 

And hence answer is -27w2

 

Hope that helps.

 

All the best.

Regards,

Ashwin (IIT Madras).

ankit singh
askIITians Faculty 614 Points
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 :)

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