Question

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

• A. $\frac{a+b+c}{3}$
• B. $\frac{a+{\omega }^{2}b+\omega c}{3}$
• C. $\frac{a+\omega b+{\omega }^{2}c}{3}$
• D. $\frac{a+b+c\omega }{3}$

Answer 1

The correct option is A $\frac{a+b+c}{3}$

Given $\omega$ is a cube root of unity and $x+y+z=a,x+\omega y+{\omega }^{2}z=b,x+{\omega }^{2}y+\omega z=c$.

Adding all the given equations gives

$3x+y(1+\omega +{\omega }^2)+z(1+\omega +{\omega }^2)=a+b+c$

$\Rightarrow 3x=a+b+c$ $\because 1+\omega +{\omega }^2=0$

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

Hence, option A.