Question

The value of $\frac{\left(a+b\omega +c{\omega }^2\right)}{\left(b+c\omega +a{\omega }^2\right)}+\frac{\left(a+b\omega +c{\omega }^2\right)}{\left(c+a\omega +b{\omega }^2\right)}$ will be

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is B

$-1$

Explanation for the correct option:

Find the value of the given expression.

We know that, $\omega$ is the cube root of unity and $1+\omega +{\omega }^2=0$ and $\omega ={\omega }^4$ also, ${\omega }^3=1$.

Simplify the given expression as follows:

$$\begin{gathered} \frac{\left(a+b\omega +c{\omega }^2\right)}{\left(b+c\omega +a{\omega }^2\right)}+\frac{\left(a+b\omega +c{\omega }^2\right)}{\left(c+a\omega +b{\omega }^2\right)}=\frac{{\omega }^2\left(a+b\omega +c{\omega }^2\right)}{{\omega }^2\left(b+c\omega +a{\omega }^2\right)}+\frac{{\omega }^2\left(a+b\omega +c{\omega }^2\right)}{{\omega }^2\left(c+a\omega +b{\omega }^2\right)} \\ \Rightarrow =\frac{\left(a{\omega }^2+b{\omega }^3+c{\omega }^4\right)}{{\omega }^2\left(b+c\omega +a{\omega }^2\right)}+\frac{{\omega }^2\left(a+b\omega +c{\omega }^2\right)}{\left(c{\omega }^2+a{\omega }^3+b{\omega }^4\right)} \\ \Rightarrow =\frac{\left(a{\omega }^2+b+c{\omega }^3\omega \right)}{{\omega }^2\left(b+c\omega +a{\omega }^2\right)}+\frac{{\omega }^2\left(a+b\omega +c{\omega }^2\right)}{\left(c{\omega }^2+a+b{\omega }^3\omega \right)} \\ \Rightarrow =\frac{\left(b+c\omega +a{\omega }^2\right)}{{\omega }^2\left(b+c\omega +a{\omega }^2\right)}+\frac{{\omega }^2\left(a+b\omega +c{\omega }^2\right)}{\left(a+b\omega +c{\omega }^2\right)} \\ \Rightarrow =\frac{1}{{\omega }^2}+\frac{{\omega }^2}{1} \\ \Rightarrow =\frac{1+{\omega }^4}{{\omega }^2} \\ \Rightarrow =\frac{1+\omega }{{\omega }^2} \\ \Rightarrow =\frac{-{\omega }^2}{{\omega }^2} \\ \Rightarrow =-1 \end{gathered}$${Since, $1+\omega =-{\omega }^2$}

Therefore, the value of the given expression is $-1$.

Hence, option (B) is the correct answer.