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

ω, ω2 are roots of 1a+x + 1b+x + 1c+x =2xPut x = 1⇒ 1a+1 + 1b+1 + 1c+1 = 2

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

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answer is 2.

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Detailed Solution

$ω, ω^{2}$ are roots of $\frac{1}{a + x} + \frac{1}{b + x} + \frac{1}{c + x} = \frac{2}{x}$

Put $x = 1 \Rightarrow \frac{1}{a + 1} + \frac{1}{b + 1} + \frac{1}{c + 1} = 2$

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