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ifomegaa n domega^2 are the nonreal cube...

if`omegaa n domega^2` are the nonreal cube roots of unity and `[1//(a+omega)]+[1//(b+omega)]+[1//(c+omega)]=2omega^2` and `[1//(a+omega)^2]+[1//(b+omega)^2]+[1//(c+omega)^2]=2omega^` , then find the value of `[1//(a+1)]+[1//(b+1)]+[1//(c+1)]dot`

Text Solution

Verified by Experts

The given relation can be rewritten as
`(1)/(a+omega) + (1)/(b+omega) + (1)/(c+omega) = (2)/(omega)`
and `(1)/(a+omega^(2)) + (1)/(b+omega^(2)) + (1)/(c+ omega^(2)) = (2)/(omega^(2))`
`rArr omega and omega^(2)` are roots of `(1)/(a+x) +(1)/(b+omega^(2)) +(1)/(c+x) = (2)/(x)`
Tow roots of the (1) are `omega` and `omega^(2)`.Let the third root be `alpha`.Then, `alpha +omega+ omega^(2) =0 or aklpha = - omega -omega^(2) = 1`
Therefore `alpha = 1` will satisfy Eq.(1)
Hence, `(1)/(a+1)+(1)/(b+1)+(1)/(c+1) = 2`
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