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(D) Given that $a = e^{i\alpha}$,$b = e^{i\beta}$,and $c = e^{i\gamma}$.Substituting these into the equation $\frac{b}{c} + \frac{c}{a} + \frac{a}{b}

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(D) Given that $a = e^{i\alpha}$,$b = e^{i\beta}$,and $c = e^{i\gamma}$.Substituting these into the equation $\frac{b}{c} + \frac{c}{a} + \frac{a}{b} = 1$,we get:$e^{i(\beta - \gamma)} + e^{i(\gamma - \alpha)} + e^{i(\alpha - \beta)} = 1$.Using Euler's formula $e^{i	heta} = \cos 	heta + i\sin 	heta$,we have:$(\cos(\beta - \gamma) + i\sin(\beta - \gamma)) + (\cos(\gamma - \alpha) + i\sin(\gamma - \alpha)) + (\cos(\alpha - \beta) + i\sin(\alpha - \beta)) = 1 + 0i$.Equating the real parts on both sides:$\cos(\beta - \gamma) + \cos(\gamma - \alpha) + \cos(\alpha - \beta) = 1$.

If $a = \cos \alpha + i\sin \alpha$,$b = \cos \beta + i\sin \beta$,$c = \cos \gamma + i\sin \gamma$ and $\frac{b}{c} + \frac{c}{a} + \frac{a}{b} = 1$,then $\cos (\beta - \gamma) + \cos (\gamma - \alpha) + \cos (\alpha - \beta)$ is equal to

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