Question

If cube root of unity is $1,\omega ,{\omega }^2$ then find 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. $\omega$
• B. $1$
• C. $-\omega$
• D. $-1$

Answer 1

The correct option is D $-1$

$\left(\frac{\omega }{\omega }\right)\times \left(\frac{a+b\omega +c{\omega }^2}{c+a\omega +b{\omega }^2}\right)+\left(\frac{a+b\omega +c{\omega }^2}{b+c\omega +a{\omega }^2}\right)\times \left(\frac{{\omega }^2}{{\omega }^2}\right)=\left(\frac{a\omega +b{\omega }^2+c{\omega }^3}{\omega (c+a\omega +b{\omega }^2)}\right)+\left(\frac{a{\omega }^2+b{\omega }^3+c{\omega }^4}{(b+c\omega +a{\omega }^2)\times {\omega }^2}\right)=\left(\frac{a\omega +b{\omega }^2+c}{\omega (c+a\omega +b{\omega }^2)}\right)+\left(\frac{a{\omega }^2+b+c\omega }{(b+c\omega +a{\omega }^2){\omega }^2}\right)=\left(\frac{1}{\omega }\right)+\left(\frac{1}{{\omega }^2}\right)=\left(\frac{{\omega }^2+\omega }{{\omega }^3}\right)=\left(\frac{-1}{1}\right)=-1$