Question

If $1,\omega ,{\omega }^2$ are cube roots of unity, then the value of $\frac{a\omega +b{\omega }^2+c{\omega }^3+d\omega }{c\omega +d{\omega }^2+a{\omega }^2+b}$ equals

• A. $\omega$
• B. ${\omega }^2$
• C. $0$
• D. None of these

Answer 1

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

$\frac{aw+b{w}^2+c{w}^3+dw}{a{w}^2+b+cw+d{w}^2}$
Multiplying and dividing by $w$ we get
$=\frac{1}{w}\left[\frac{a{w}^2+b{w}^3+c{w}^4+d{w}^2}{a{w}^2+b+cw+d{w}^2}\right]$
$=\frac{1}{w}\left[\frac{a{w}^2+b+c{w}^3.w+d{w}^2}{a{w}^2+b+cw+d{w}^2}\right]$
$=\frac{1}{w}\left[\frac{a{w}^2+b+cw+d{w}^2}{a{w}^2+b+cw+d{w}^2}\right]$
$=\frac{1}{w}\left(1\right)$
$=\overline{w}$
$=w^2$
Hence, option 'B' is correct.