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if alpha,beta,gamma are the roots of x^3...

if `alpha,beta,gamma `are the roots of `x^3-3x^2 +3x + 7 =0` then `(alpha-1)/(beta-1)+(beta-1)/(gamma-1)+(gamma-1)/(alpha-1)`

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Verified by Experts

The correct Answer is:
`3omega^(2)`
`x^(3) -3x +3x +7=0`
`rArr (x-1)^(3)= - 8`
`rArr (x-1)/(2) =(1)^(1//3) =1, omega,omega^(2)`
`rArr alpha = - 1, beta = 1 - 2 omega, gamma= 1 - 2 omega^(2)`
`therefore " " E= (-2)/(-2Omega)+(-2omega)/(-2omega^(2))+(-2omega^(2))/(-2)`
`therefore" "E=(-2)/(-2omega)+(-2omega)/(-2omega^(2))+(-2Omega^(2))/(-2)`
`=(1)/(omega)+(1)/(Omega)+(1)/(omega)+(3)/(omega)=3omega^(2)`
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