If 1, omega, omega^2 are the cube roots of un | vedclass

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(D) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2 = 0$ and $\omega^3 = 1$. We need to evaluate the expr

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(D) Given that $1, \omega, \omega^2$ are the cube roots of unity,we know that $1+\omega+\omega^2 = 0$ and $\omega^3 = 1$. We need to evaluate the expression: $\frac{1}{1+2 \omega}+\frac{1}{2+\omega}-\frac{1}{1+\omega}$. First,simplify the first two terms: $\frac{1}{1+2 \omega}+\frac{1}{2+\omega} = \frac{2+\omega+1+2 \omega}{(1+2 \omega)(2+\omega)} = \frac{3+3 \omega}{2+\omega+4 \omega+2 \omega^2} = \frac{3(1+\omega)}{2+5 \omega+2 \omega^2}$. Since $1+\omega = -\omega^2$,the numerator is $-3 \omega^2$. Also,$2+5 \omega+2 \omega^2 = 2(1+\omega^2)+5 \omega = 2(-\omega)+5 \omega = 3 \omega$. So,the first two terms sum to $\frac{3(1+\omega)}{3 \omega} = \frac{1+\omega}{\omega} = \frac{-\omega^2}{\omega} = -\omega$. Now,subtract the third term: $-\frac{1}{1+\omega} = -\frac{1}{-\omega^2} = \frac{1}{\omega^2} = \omega$. Thus,the total sum is $-\omega + \omega = 0$.

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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