If $\omega$ is the cube root of unity $\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b…

If $\omega$ is the cube root of unity $\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2}=$
  1. 2
  2. -2
  3. 1
  4. -1

Solution

$\begin{aligned} & \text { } \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{a+b \omega+c \omega^2}{b+c \omega+a \omega^2} \\ & \frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}+\frac{\omega^2\left(a+b \omega+c \omega^2\right)}{b \omega^2+c+a \omega} \\ & =\left(1+\omega^2\right)\left(\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}\right)=-\frac{\left(a \omega+b \omega^2+c\right)}{\left(c+a \omega+b \omega^2\right)}=-1\end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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