Question

$\frac{a+b\omega +c{\omega }^2}{c+a\omega +b{w}^2}=$

• A. $0$
• B. $1$
• C. $\omega$
• D. ${\omega }^2$

Answer 1

The correct option is D ${\omega }^2$

Multiply & divide the given equation by $\omega ,$

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

$=\frac{1}{\omega }$

$=\frac{1}{\omega }\times \frac{{\omega }^2}{{\omega }^2}$

$=\frac{{\omega }^2}{{\omega }^3}$

$=\frac{{\omega }^2}{1}$

$={\omega }^2$