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

• A. 1
• B. -1
• C. 2
• D. -2

Answer 1

The correct option is B -1

Given, $\frac{a+b\omega +c{\omega }^2}{b+c\omega +a{\omega }^2}+\frac{a+b\omega +c{\omega }^2}{c+a\omega +b{\omega }^2}$
$=\frac{a+b\omega +c{\omega }^2}{b+c\omega +a{\omega }^2}+\frac{\omega }{\omega }\cdot \frac{a+b\omega +c{\omega }^2}{c+a\omega +b{\omega }^2}$
$=\frac{a+b\omega +c{\omega }^2}{b+c\omega +a{\omega }^2}+\frac{a\omega +b{\omega }^2+c{\omega }^3}{c\omega +a{\omega }^2+b{\omega }^3}$
$=\frac{a+b\omega +c{\omega }^2}{b+c\omega +a{\omega }^2}+\frac{a\omega +b{\omega }^2+c}{c\omega +a{\omega }^2+b},[\because {\omega }^3=1]$
$=\frac{a+b\omega +c{\omega }^2+a\omega +b{\omega }^2+c}{b+c\omega +a{\omega }^2}$
$=\frac{a(1+\omega )+b(\omega +{\omega }^2)+c(1+{\omega }^2)}{b+c\omega +a{\omega }^2}$
$=\frac{a(-{\omega }^2)+b(-1)+c(-\omega )}{b+c\omega +a{\omega }^2},[\because 1+\omega +{\omega }^2=0]$
$=-1$