Complex Numbers And Quadratic Equations question 250
Question: 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}}} $ will be [BIT Ranchi 1989; Orissa JEE 2003]
Options:
A) 1
B) - 1
C) 2
D) - 2
Show Answer
Answer:
Correct Answer: B
Solution:
Multiplying the numerator and denominator by $ \omega $ and $ {{\omega }^{2}} $ respectively I and II expressions $ =\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}}} $ $ =\frac{\omega (a+b\omega +c{{\omega }^{2}})}{(b\omega +c{{\omega }^{2}}+a)}+\frac{{{\omega }^{2}}(a+b\omega +c{{\omega }^{2}})}{(c{{\omega }^{2}}+a+a\omega )}=\omega +{{\omega }^{2}}=-1 $
BETA
