Question

If $x=a+b$, $y=a\omega +b{\omega }^2$ and $z=a{\omega }^2+b\omega$ where $w$ is complex cube root of unity
then (i) $x+y+z=0$
(ii) $xyz=a^3+b^3$

• A. True
• B. False

Answer 1

The correct option is A True

Given:$x=a+b,y=a\omega +b{\omega }^2,y=a\omega +b{\omega }^2$ and $z=a{\omega }^2+b\omega$

$\left(i\right)x+y+z=a+b+a\omega +b{\omega }^2+a{\omega }^2+b\omega$

$=a(1+\omega +{\omega }^2)+b(1+\omega +{\omega }^2)$

$=a\times 0+b\times 0$ since $1+\omega +{\omega }^2=0$

$=0$

$\left(ii\right)xyz=(a+b)(a\omega +b{\omega }^2)(a{\omega }^2+b\omega )$

$=(a+b)(a^2{\omega }^3+ab{\omega }^2+ab{\omega }^4+b^2{\omega }^3)$

$=(a+b){\omega }^3(a^2+ab{\omega }^2+ab\omega +b^2)$

$=(a+b)(a^2+ab({\omega }^2+\omega )+b^2)$

$=(a+b)(a^2+ab(-1)+b^2)$ since $1+\omega +{\omega }^2=0$

$=(a+b)(a^2-ab+b^2)$

$=a^3+b^3$

Hence the above statements are true.