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(C) Let $X = a+b\omega+c\omega^2$. Note that $\omega X = a\omega+b\omega^2+c$ and $\omega^2 X = a\omega^2+b+c\omega$.The given expression is $\left(\f

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divisible by $3$

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(C) Let $X = a+b\omega+c\omega^2$. Note that $\omega X = a\omega+b\omega^2+c$ and $\omega^2 X = a\omega^2+b+c\omega$.The given expression is $\left(\frac{X}{c+a\omega+b\omega^2}\right)^k + \left(\frac{X}{b+a\omega^2+c\omega}\right)^l = 2$.Since $c+a\omega+b\omega^2 = \omega X$ and $b+a\omega^2+c\omega = \omega^2 X$,we substitute these:$\left(\frac{X}{\omega X}\right)^k + \left(\frac{X}{\omega^2 X}\right)^l = 2$$\left(\frac{1}{\omega}\right)^k + \left(\frac{1}{\omega^2}\right)^l = 2$$\omega^{-k} + \omega^{-2l} = 2$Since $\omega^3 = 1$,we have $\omega^{-k} = \omega^{3n-k}$ and $\omega^{-2l} = \omega^{3m-2l}$.For the sum of two complex numbers of modulus $1$ to be $2$,both must be $1$.Thus,$\omega^{-k} = 1$ and $\omega^{-2l} = 1$,which implies $k$ is a multiple of $3$ and $2l$ is a multiple of $3$ (so $l$ is a multiple of $3$).Therefore,$2k+l$ must be a multiple of $3$.

If $\omega$ is the complex cube root of unity and $\left(\frac{a+b \omega+c \omega^2}{c+a \omega+b \omega^2}\right)^k+\left(\frac{a+b \omega+c \omega^2}{b+a \omega^2+c \omega}\right)^l=2$,then $2k+l$ is always

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