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If $ω \neq 1$ is a cube root of unity and satisfies
$\frac{1}{a + ω} + \frac{1}{b + ω} + \frac{1}{c + ω} = 2 ω^{2}$
and $\frac{1}{a + ω^{2}} + \frac{1}{b + ω^{2}} + \frac{1}{c + ω^{2}} = 2 ω,$
then the value of $\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1}$ then

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-2

=1+ω2

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Correct option is A

Note that ω , ω2are roots of 1a+x+1b+x+1c+x=2x⇔xbc+ca+ab+2(a+b+c)x+3x2=2abc+(bc+ca+ab)x+(a+b+c)x2+x3If α is the third root of this equation, then α+ω+ω2=0⇒α=1

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If ω≠1 is a cube root of unity and satisfies 1a+ω+1b+ω+1c+ω=2ω2and 1a+ω2+1b+ω2+1c+ω2=2ω,then the value of 1a+1+1b+1+1c+1 then

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If $ω \neq 1$ is a cube root of unity and satisfies
$\frac{1}{a + ω} + \frac{1}{b + ω} + \frac{1}{c + ω} = 2 ω^{2}$
and $\frac{1}{a + ω^{2}} + \frac{1}{b + ω^{2}} + \frac{1}{c + ω^{2}} = 2 ω,$
then the value of $\frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1}$ then