Question

If $x=a+b,y=a\omega +b{\omega }^2$ and $z=a{\omega }^2+b\omega$, where $\omega$ and ${\omega }^2$ are cube show that $x^2+y^2+z^2=6ab$.

Answer 1

$x=a+b$

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

$z=a{\omega }^2+b\omega$

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

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

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

$=6ab+a^2(1+\omega +{\omega }^2)+b^2(1+\omega +{\omega }^2)$

$=6ab+a^2\left(0\right)+b^2\left(0\right)$

$=6ab$

Hence showed.