Q. If $\omega$ is a complex cube root of unity, then $\frac{a + b\omega + c\omega^{2}}{c + a\omega + b\omega^{2}} + \frac{c + a\omega + b\omega^{2}}{a + b\omega + c\omega^{2}} + \frac{b + c\omega + a\omega^{2}}{b + c\omega^{4} + a\omega^{5}}$ is equal to
A) 1
B) $\omega$
C) $\omega^{2}$
D) 0
Explanation
(d)
$\frac{a + b\omega + c\omega^{2}}{c + a\omega + b\omega^{2}} + \frac{c + a\omega + b\omega^{2}}{a + b\omega + c\omega^{2}} + \frac{b + c\omega + a\omega^{2}}{b + c\omega^{4} + a\omega^{5}}$
$= \frac{\omega^{2}(a + b\omega + c\omega^{2})}{(a + b\omega + c\omega^{2})} + \frac{\omega(a\omega + b\omega^{2} + c)}{(a\omega + b\omega^{2} + c)} + \frac{(b + c\omega + a\omega^{2})}{(b + c\omega + a\omega^{2})}$
$= \omega^{2} + \omega + 1 = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.