If omega is a complex cube root of unity and | vedclass

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(B) Let $S = \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} + \frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}$.Note that $c+a \omega+b \omega

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) Let $S = \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2} + \frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}$.Note that $c+a \omega+b \omega^2 = \omega^2(c \omega + a \omega^2 + b) = \omega^2(b+c \omega+a \omega^2)$.Thus,the expression becomes $\frac{a+b \omega+c \omega^2}{\omega^2(b+c \omega+a \omega^2)} + \frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}$.Factoring out $\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}$,we get $\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2} (\frac{1}{\omega^2} + 1) = \frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2} (\omega + 1)$.Since $1+\omega+\omega^2 = 0$,we have $1+\omega = -\omega^2$.However,evaluating the expression for specific values like $a=1, b=0, c=0$ gives $\frac{1}{\omega^2} + \frac{1}{\omega} = \omega + \omega^2 = -1$.Thus,the value is $-1$.

If $\omega$ is a complex cube root of unity and $a, b, c$ are distinct real numbers,then $\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} = $

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