If 1, ω, ω^2 are the cube roots of unity, then 1/(1+2 ω)+1/(2+ω)-1/(1+ω)=

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MARKS Content Team · Accepted Answer

Correct answer: (D) 0. Given, 1/(1+2 ω)+1/(2+ω)-1/(1+ω) =(2+ω+1+2 ω)/((1+2 ω)(2+ω))-1/((1+ω)) =(3+3 ω)/((1+2 ω)(2+ω))-1/((1+ω)) =((3+3 ω)(1+ω)-(1+2 ω)(2+ω))/((1+2 ω)(2+ω)(1+ω)) =(3+3 ω+3 ω+3 ω^2- (2+ω+4 ω+2 ω^2 ))/((1+2 ω)(2+ω)(1+ω)) =(3+6 ω+3 ω^2-2-w-4 ω-2 ω^2)/((1+2 ω)(2+ω)(1+ω)) =(1+ω+ω^2)/((1+2 ω)(2+ω)(1+ω)) =0

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

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