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If p=a+bomega+comega^2;q=b+comega+aomega...

If `p=a+bomega+comega^2;q=b+comega+aomega^2` and `r=c+aomega+bomega^2` where `a,b,c!=0` and `omega` is the complex cube root of unity, then

A

`p+q+r=a+b+c`

B

`p^2+q^2+r^2=a^2+b^2+^2c`

C

`p^2+q^2+r^2=-2(pq+qr+rp)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
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