Question

If $1,\omega ,{\omega }^2$ are cube roots of unity and $x=a+b$, $y=a{\omega }^2+b\omega ,z=a\omega +b{\omega }^2$ , then $x^2+y^2+z^2=$

• A. $6ab$
• B. $12ab$
• C. $8a^2{b}^2$
• D. $4a^2{b}^2$

Answer 1

The correct option is A $6ab$

$x^2=(a+b{)}^2=a^2+b^2+2ab$ ...... $\left(1\right)$
$y^2=(a{\omega }^2+b\omega {)}^2=a^2{\omega }^4+b^2{\omega }^2+2ab{\omega }^3$ ...... $|\because {\omega }^3=1|$
$=a^2\omega +b^2{\omega }^2+2ab$ ....... $\left(2\right)$
$z^2=(a\omega +b{\omega }^2{)}^2=a^2{\omega }^2+b^2{\omega }^4+2ab{\omega }^3$
$=a^2{\omega }^2+b^2\omega +2ab$ ..... $\left(3\right)$
$x^2+y^2+z^2=a^2(1+\omega +{\omega }^2)+b^2(1+\omega +{\omega }^2)+6ab$
$=6ab[\because 1+\omega +{\omega }^2=0]$