The value of frac{a + bomega + comega^2}{b + | vedclass

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(B) Let the given expression be $E = \frac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2} + \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2}

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) Let the given expression be $E = \frac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2} + \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2}$.Multiply the numerator and denominator of the first term by $\omega$:$\frac{\omega(a + b\omega + c\omega^2)}{\omega(b + c\omega + a\omega^2)} = \frac{\omega(a + b\omega + c\omega^2)}{b\omega + c\omega^2 + a\omega^3} = \frac{\omega(a + b\omega + c\omega^2)}{a + b\omega + c\omega^2} = \omega$.Multiply the numerator and denominator of the second term by $\omega^2$:$\frac{\omega^2(a + b\omega + c\omega^2)}{\omega^2(c + a\omega + b\omega^2)} = \frac{\omega^2(a + b\omega + c\omega^2)}{c\omega^2 + a\omega^3 + b\omega^4} = \frac{\omega^2(a + b\omega + c\omega^2)}{c\omega^2 + a + b\omega} = \omega^2$.Thus,$E = \omega + \omega^2$.Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.

The value of $\frac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2} + \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2}$ is

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