Question

If $\omega$ is an imaginary cube root of unity and $x=a+b,y=a\omega +b{\omega }^2,z=a{\omega }^2+b\omega$, then $x^2+y^2+z^2$ is equal to

• A. $6ab$
• B. $3ab$
• C. $6a^{2b^2}$
• D. $3a^{2b^2}$

Answer 1

The correct option is A

$6ab$

Explanation for the correct option

Step 1: Prerequisites for the solution

When $\omega$ is an imaginary cube root of unity then we know that

${\omega }^3=1\text{and}1+\omega +{\omega }^2=0$

Step 2: Simplification and calculation for the solution

As $x=a+b$ then,

$\begin{array}{rcl}{x}^2& =& {\left(a+b\right)}^2\\ & =& a^2+b^2+2ab\end{array}$

Let us call it equation (1)

Now $y=a\omega +b{\omega }^2$ then

$\begin{array}{rcl}{y}^2& =& {\left(a\omega +b{\omega }^2\right)}^2\\ & =& a^2{\omega }^2+b^2{\omega }^4+2ab{\omega }^3\\ & =& a^2{\omega }^2+b^2\omega +2ab\end{array}$

This will be the equation (2)

Also, $z=a{\omega }^2+b\omega$ implies that

$\begin{array}{rcl}{z}^2& =& {\left(a{\omega }^2+b\omega \right)}^2\\ & =& a^2{\omega }^4+b^2{\omega }^2+2ab{\omega }^3\\ & =& a^2\omega +b^2{\omega }^2+2ab\end{array}$

This will be the equation (3)

Now from equations (1), (2), and (3) we have

$\begin{array}{rcl}{x}^2+y^2+z^2& =& a^2+b^2+2ab+a^2{\omega }^2+b^2\omega +2ab+a^2\omega +b^2{\omega }^2+2ab\\ & =& a^2\left(1+{\omega }^2+\omega \right)+b^2\left(1+\omega +{\omega }^2\right)+6ab\end{array}$

Since $1+\omega +{\omega }^2=0$ then

$x^2+y^2+z^2=6ab$

Hence, the correct option is (A).