Question

If $\omega$ is cube root of unity and $x+y+z=a$, $x+\omega y+{\omega }^{2}z=b$, $x+{\omega }^{2}y+\omega z=b$ then which of the following is not correct?

• A. $x=\frac{a+b+c}{3}$
• B. $y=\frac{a+b{\omega }^2+\omega c}{3}$
• C. $x=\frac{a+b\omega +{\omega }^{2}c}{3}$
• D. None of these

Answer 1

The correct option is D None of these

Given $x+y+z=a$...(i)

$x+\omega y+{\omega }^{2}z=b$ ...(ii)

$x+{\omega }^{2}y+\omega z=c$ ...(iii)

By adding (i), (ii) and (iii), we get

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

Hence option A is correct.
Again $\left(i\right)+\left(ii\right)\times {\omega }^2+\left(iii\right)\times \omega$, we get

$3y=a{\omega }^3+b{\omega }^2+c\omega$

$y=\frac{a+b{\omega }^2+c\omega }{3}$

Hence, option B is correct.
Similarly, $\left(i\right)+\left(ii\right)\times \omega +\left(iii\right)\times {\omega }^2$ we get

$z=\frac{a+b\omega +c{\omega }^2}{3}$

Hence, option C is correct