If omega is a complex cube root of unity,then | vedclass

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(D) Let the given expression be $E = \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}$.We know

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$-1$

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(D) Let the given expression be $E = \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}$.We know that $\omega^3 = 1$ and $1+\omega+\omega^2 = 0$.Multiply the numerator and denominator of the second term by $\omega$:$E = \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} + \frac{\omega(a+b \omega+c \omega^2)}{\omega(b+c \omega+a \omega^2)}$$E = \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} + \frac{a\omega+b \omega^2+c \omega^3}{b\omega+c \omega^2+a \omega^3}$Since $\omega^3 = 1$,the second term becomes $\frac{a\omega+b \omega^2+c}{b\omega+c \omega^2+a}$.This is equal to $\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} 	imes \omega$ is not correct,let's re-evaluate:Note that $\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} 	imes \omega = \frac{a\omega+b \omega^2+c \omega^3}{c\omega+a \omega^2+b} = \frac{a\omega+b \omega^2+c}{b+c\omega+a \omega^2}$.Thus,$E = \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} + \omega \left( \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} ight) = (1+\omega) \left( \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} ight)$.Using $1+\omega = -\omega^2$,we get $E = -\omega^2 \left( \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} ight) = \frac{-a\omega^2-b\omega^3-c\omega^4}{c+a\omega+b\omega^2} = \frac{-a\omega^2-b-c\omega}{c+a\omega+b\omega^2} = -1$.

If $\omega$ is a complex cube root of unity,then the value of $\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}$ is:

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