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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Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { Given, } \frac{1}{1+2 \omega}+\frac{1}{2+\omega}-\frac{1}{1+\omega} \ & =\frac{2+\omega+1+2 \omega}{(1+2 \omega)(2+\omega)}-\frac{1}{(1+\omega)} \ & =\frac{3+3 \omega}{(1+2 \omega)(2+\omega)}-\frac{1}{(1+\omega)} \ & =\frac{(3+3 \omega)(1+\omega)-(1+2 \omega)(2+\omega)}{(1+2 \omega)(2+\omega)(1+\omega)} \ & =\frac{3+3 \omega+3 \omega+3 \omega^2-\left(2+\omega+4 \omega+2 \omega^2ight)}{(1+2 \omega)(2+\omega)(1+\omega)} \ & =\frac{3+6 \omega+3 \omega^2-2-w-4 \omega-2 \omega^2}{(1+2 \omega)(2+\omega)(1+\omega)} \ & =\frac{1+\omega+\omega^2}{(1+2 \omega)(2+\omega)(1+\omega)} \ & =0 \end{aligned} $

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

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