Question

If we are given that $\omega$ is the 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 not correct?
(a) $x=\dfrac{a+b+c}{3}$
(b) $y=\dfrac{a+b{{\omega }^{2}}+\omega c}{3}$
(c) $x=\dfrac{a+b\omega +{{\omega }^{2}}c}{3}$
(d) None of these

Answer 1

To solve most of the questions on the cube root of unity, you must know its following properties:

1. Sum of all three cube roots of unity is zero i.e. $1+\omega +{{\omega }^{2}}=0$
2. Product of all three cube roots of unity is one i.e. ${{\omega }^{3}}=1$
   To get started, we have to add all the given equations and use property (1) mentioned above to get the value of x. Again we have to perform operations like multiplying the second equation with ${{\omega }^{2}}$ and the third equation with $\omega$ and then applying property (2) to simplify them. Obtain values of y and z too and then choose the right option.

Complete step-by-step solution:
Let us assume that
$x+y+z=a\text{ }..............\text{ (1)}$
$x+\omega y+{{\omega }^{2}}z=b\text{ }..............\text{ (2)}$
$x+{{\omega }^{2}}y+\omega z=c\text{ }..............\text{ (3)}$
Adding equation (1), (2) and (3), we get
$3x+(1+\omega +{{\omega }^{2}})y+(1+\omega +{{\omega }^{2}})z=a+b+c$
As we know that $1+\omega +{{\omega }^{2}}=0$, we get
$\begin{aligned} & \Rightarrow 3x=a+b+c \\\ & \Rightarrow x=\dfrac{a+b+c}{3} \\\ \end{aligned}$
Now multiplying equation (2) with ${{\omega }^{2}}$ , we get

& {{\omega }^{2}}x+{{\omega }^{3}}y+{{\omega }^{4}}z=b{{\omega }^{2}} \\\ & \Rightarrow {{\omega }^{2}}x+{{\omega }^{3}}y+\omega .{{\omega }^{3}}z=b{{\omega }^{2}} \\\ \end{aligned}$$ As we know that ${{\omega }^{3}}=1$ $$\Rightarrow {{\omega }^{2}}x+y+\omega z=b{{\omega }^{2}}\text{ }..............\text{ (4)}$$ Now multiplying equation (3) with $\omega $ , we get $$\omega x+{{\omega }^{3}}y+{{\omega }^{2}}z=c\omega $$ As we know that ${{\omega }^{3}}=1$ $$\Rightarrow \omega x+y+{{\omega }^{2}}z=c\omega \text{ }..............\text{ (5)}$$ Adding equation (1), (4) and (5), we get $(1+\omega +{{\omega }^{2}})x+3y+(1+\omega +{{\omega }^{2}})z=a+b{{\omega }^{2}}+c\omega $ As we know that $1+\omega +{{\omega }^{2}}=0$ $\begin{aligned} & \Rightarrow 3y=a+b{{\omega }^{2}}+c\omega \\\ & \Rightarrow y=\dfrac{a+b{{\omega }^{2}}+c\omega }{3} \\\ \end{aligned}$ Now multiplying equation (2) with $\omega $ , we get $$\omega x+{{\omega }^{2}}y+{{\omega }^{3}}z=b\omega $$ As we know that ${{\omega }^{3}}=1$ $$\Rightarrow \omega x+{{\omega }^{2}}y+z=b\omega \text{ }..............\text{ (6)}$$ Now multiplying equation (3) with ${{\omega }^{2}}$ , we get $$\begin{aligned} & {{\omega }^{2}}x+{{\omega }^{4}}y+{{\omega }^{3}}z=c{{\omega }^{2}} \\\ & \Rightarrow {{\omega }^{2}}x+\omega .{{\omega }^{3}}y+{{\omega }^{3}}z=c{{\omega }^{2}} \\\ \end{aligned}$$ As we know that ${{\omega }^{3}}=1$ $$\Rightarrow {{\omega }^{2}}x+\omega y+z=c{{\omega }^{2}}\text{ }..............\text{ (7)}$$ Adding equation (1), (6) and (7), we get $(1+\omega +{{\omega }^{2}})x+(1+\omega +{{\omega }^{2}})y+3z=a+b\omega +c{{\omega }^{2}}$ As we know that $1+\omega +{{\omega }^{2}}=0$ $\begin{aligned} & \Rightarrow 3z=a+b\omega +c{{\omega }^{2}} \\\ & \Rightarrow z=\dfrac{a+b\omega +c{{\omega }^{2}}}{3} \\\ \end{aligned}$ So, we got $x=\dfrac{a+b+c}{3}$ , $y=\dfrac{a+b{{\omega }^{2}}+c\omega }{3}$ and $z=\dfrac{a+b\omega +c{{\omega }^{2}}}{3}$ **Hence, Option (c) is correct as it doesn’t match with the results that we have got.** **Note:** The key point that we used here is that the coefficient of the variable whose value we need to find should be made unity by multiplying it with the appropriate cube root of unity and the rest of the variables are eliminated by using the properties of the cube root of unity. These type of questions checks how familiar you are with the properties and if you try to do it without applying properties then you’ll make things complicated, so always try to use the properties whenever you see cube root of unity in a question.