Question

If $\omega$ is a complex cube root of unity, the the value of $\frac{a+b\omega +c{\omega }^2}{c+a\omega +b{\omega }^2}+\frac{a+b\omega +c{\omega }^2}{b+c\omega +a{\omega }^2}$

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

Answer: (B)

$-1$

$\omega$ us complex root of unity $\Rightarrow \omega =1$and $1+\omega +{\omega }^2=0$

$\frac{a+b\omega +C{\omega }^2}{C+a\omega +b{\omega }^2}+\frac{a+b\omega +C{\omega }^2}{b+C\omega +a{\omega }^2}$

$\frac{(a+b\omega +c{\omega }^2)\omega }{(c+a\omega +b{\omega }^2)\omega }+\frac{(a+b\omega +c{\omega }^2)}{(b+c\omega +a{\omega }^2)}$

$\frac{(a\omega +b{\omega }^2+c)}{c\omega +a{\omega }^2+b}+\frac{a+b\omega +c{\omega }^2)}{(b+c\omega +a{\omega }^2)}$ $(\because {\omega }^3\equiv 1)$

$\frac{a(1+\omega )+b(\omega +{\omega }^2)+c(1+{\omega }^2)}{(b+c\omega +a{\omega }^2)}$

$\frac{[b+c\omega +a{\omega }^2]}{(b+c\omega +a{\omega }^2)}$

$-1$

$\therefore \frac{a+b\omega +c{\omega }^2}{c+a\omega +b{\omega }^2}+\frac{a+b\omega +c{\omega }^2}{b+c\omega +a{\omega }^2}=-1$