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If $\omega$ is the cube root of unity then the value of $\large\frac{a\omega+b+c\omega^2}{a\omega^2+b\omega+c}+\frac{a\omega^2+b+c\omega}{a+b\omega+c\omega^2}=?$

(A) $\omega$ (B) $\omega^2$ (C) $2\omega$ (D) $2\omega^2$

1 Answer

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  • $\omega^3=1$
$\large\frac{a\omega+b+c\omega^2}{a\omega^2+b\omega+c}$$=\large\frac{a\omega+b+c\omega^2}{a\omega^2+b\omega+c}\times\frac{\omega}{\omega}$
$=\large\frac{1}{\omega}=\frac{\omega^2}{\omega^3}$$=\omega^2$
$\large\frac{a\omega^2+b+c\omega}{a+b\omega+c\omega^2}$$=\large\frac{a\omega^2+b+c\omega}{a+b\omega+c\omega^2}\times\frac{\omega^2}{\omega^2}$
$=\omega^2$
$\Rightarrow\:\large\frac{a\omega+b+c\omega^2}{a\omega^2+b\omega+c}$$+\large\frac{a\omega^2+b+c\omega}{a+b\omega+c\omega^2}$$=\omega^2+\omega^2=2\omega^2$
answered Jul 12, 2013 by rvidyagovindarajan_1
 

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