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Find the value of \( \frac{a+b \omega+c \omega^{2}}{c+a \omega+b \omega^{2}} \) in terms of \( w^{2} \) |
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Answer» \(\frac{a+b\omega+c\omega^2}{c+a\omega+b\omega^2}\) = \(\frac{a+b/\omega^2+c\omega^2}{c+a/\omega^2+b\omega^2}\) (\(\because\omega^3=1, \omega=1/\omega^2\)) = \(\frac{a\omega^2+b+c\omega^4}{c\omega^2+a+b\omega^4}\)\(=\frac{b+\omega^2(a+c\omega^2)}{a+\omega(c+b\omega^2)}\) |
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