If omega is a complex cube root of unity,then | vedclass

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(B) Let the given expression be $E = \left[\frac{51+73 \omega+87 \omega^2}{73+87 \omega+51 \omega^2}+\frac{51+73 \omega+87 \omega^2}{87+51 \omega+73 \

Vedclass · Accepted Answer

$-1$

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(B) Let the given expression be $E = \left[\frac{51+73 \omega+87 \omega^2}{73+87 \omega+51 \omega^2}+\frac{51+73 \omega+87 \omega^2}{87+51 \omega+73 \omega^2}\right]^{15}$.Multiply the numerator of the first term by $\omega$ and divide by $\omega$:$\frac{(51+73 \omega+87 \omega^2) \omega}{(73+87 \omega+51 \omega^2) \omega} = \frac{51 \omega+73 \omega^2+87 \omega^3}{73 \omega+87 \omega^2+51 \omega^3} = \frac{87+51 \omega+73 \omega^2}{73 \omega+87 \omega^2+51} = \frac{87+51 \omega+73 \omega^2}{\omega(73+87 \omega+51 \omega^2)}$.Alternatively,observe the structure:Let $A = 51+73 \omega+87 \omega^2$,$B = 73+87 \omega+51 \omega^2$,$C = 87+51 \omega+73 \omega^2$.Note that $B = \omega^2 A$ and $C = \omega A$.Thus,the expression becomes $\left[\frac{A}{\omega^2 A} + \frac{A}{\omega A}\right]^{15} = \left[\frac{1}{\omega^2} + \frac{1}{\omega}\right]^{15}$.Since $\frac{1}{\omega^2} = \omega$ and $\frac{1}{\omega} = \omega^2$,we have $(\omega + \omega^2)^{15}$.Since $1 + \omega + \omega^2 = 0$,we have $\omega + \omega^2 = -1$.Therefore,$(-1)^{15} = -1$.

If $\omega$ is a complex cube root of unity,then the value of $\left[\frac{51+73 \omega+87 \omega^2}{73+87 \omega+51 \omega^2}+\frac{51+73 \omega+87 \omega^2}{87+51 \omega+73 \omega^2}\right]^{15}$ is:

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