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Question

Find the value of the given expression:
$$\displaystyle \frac{a+b\omega +c\omega ^2}{c+a\omega +bw^2}+\frac{a+b\omega +c\omega ^2}{b+c\omega +a\omega ^2}$$


A
1
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B
0
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C
2
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D
1
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Solution

The correct option is D $$-1$$
$$\displaystyle \frac{a+bw+cw^2}{c+aw+bw^2}+\frac{a+bw+cw^2}{b+cw+aw^2}$$

$$\displaystyle =\frac{1}{w}\left \{ \frac{(aw+bw^2+cw^3)}{c+aw+bw^2} \right \}+\frac{1}{w^2}\left \{ \frac{aw^2+bw^3+cw^4}{b+cw+aw^2} \right \}$$

$$=\frac{1}{w}. 1+\frac{1}{w^2}. 1=w+w^2 \quad  \quad  \quad \begin{bmatrix}1+w+w^2=0 & \\w^3=1  & \end{bmatrix}$$

$$= - 1$$

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