Question

If ` $\omega$ ` is cube root of unity and $x+y+z=a$$x+\omega y+{\omega }^2$ $z$ $=b$, $x$ $+{\omega }^{2}y+\omega z=c$ then which of the following is correct

• A. $x=\frac{a+b+c}{3}$
• B. $y=\frac{a+b{\omega }^2+\omega c}{3}$
• C. $Z=\frac{a+b\omega +{\omega }^{2}c}{3}$
• D. All the above

Answer 1

The correct option is C $Z=\frac{a+b\omega +{\omega }^{2}c}{3}$

$\Rightarrow 1+\omega +{\omega }^2=0and{\omega }^3=1$
$x+y+z=a$
$x+wy+w^{2}z=b$
$+x+w^{2}y+wz=c$
$\overline{3x+(1+\omega +{\omega }^2)y+(1+\omega +{\omega }^2)z=a+b+c}$

$3x=a+b+cx=\frac{a+b+c}{3}$

$y+z=a-\left(\frac{a+b+c}{3}\right)=\frac{2a-b-c}{3}$

$y=\frac{2a-b-c-3z}{3}$

$\Rightarrow \frac{a+b+c}{3}+\frac{w(2a-b-c-3z)}{3}+w^{2}z=b$

$a+b+c+2aw-bw-cw-3wz+3w^{2}z=3b2b$
$a-2b+c+2aw-bw-cw=3wz-3w^{2}z$
$a(1+w)+aw-b(1+w)-b+c(1-w)=3wz(1-w)$
$-a{w}^2+aw+b{w}^2-b+c(1-w)=3wz(1-w)$
$aw(1-w)-b(1-w)(1+w)+c(1-w)=3wz(1-w)$
$aw+b{w}^2+c=3wz$

$z=\frac{w^2(aw+b{w}^2+c)}{3w^3}=\frac{a+bw+c{w}^2}{3}$
$y=\frac{2a-b-c-a-bw-c{w}^2}{3}$
$y=\frac{a-b(1+w)-c(1+w^2)}{3}$
$y=\frac{a+b{w}^2+cw}{3}$