If 1, ω, ω2 are the cube roots of unity, then 11+2ω+12+ω-11+ω=

11+2w+12+w-11+w11+2w+1+w-2-w(2+w) (1+w)11+2w+(-1)2+2w+w+w211+2w-1(1+w+w2)+(1+2w)11+2w-11+2w=0

If 1, ω, ω2 are the cube roots of unity, then 11+2ω+12+ω-11+ω=

11+2w+12+w-11+w11+2w+1+w-2-w(2+w) (1+w)11+2w+(-1)2+2w+w+w211+2w-1(1+w+w2)+(1+2w)11+2w-11+2w=0

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$If 1, ω, ω^{2} are the cube roots of unity, then \frac{1}{1 + 2 ω} + \frac{1}{2 + ω} - \frac{1}{1 + ω} =$

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$1$

$ω$

$ω^{2}$

answer is D.

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Detailed Solution

$\frac{1}{1 + 2 w} + \frac{1}{2 + w} - \frac{1}{1 + w}$

$\frac{1}{1 + 2 w} + \frac{1 + w - 2 - w}{(2 + w) (1 + w)}$

$\frac{1}{1 + 2 w} + \frac{(- 1)}{2 + 2 w + w + w^{2}}$

$\frac{1}{1 + 2 w} - \frac{1}{(1 + w + w^{2}) + (1 + 2 w)}$

$\frac{1}{1 + 2 w} - \frac{1}{1 + 2 w} = 0$

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If 1, ω, ω2 are the cube roots of unity, then 11+2ω+12+ω-11+ω=

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$If 1, ω, ω^{2} are the cube roots of unity, then \frac{1}{1 + 2 ω} + \frac{1}{2 + ω} - \frac{1}{1 + ω} =$