Question

Let $x=a+b,y=a\omega +b{\omega }^2,z=a{\omega }^2+b\omega$, where $\omega$ is the cube root of unity. Then prove than $x^2+y^2+z^2=6ab$.

Answer 1

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

Now

$x^2+y^2+z^2$

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

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

$=a^2(1+\omega +{\omega }^2)+6ab+b^2(1+{\omega }^2+\omega )$ [ Since ${\omega }^3=1$ this gives ${\omega }^4=\omega$]

$=6ab$. [ As $1+\omega +{\omega }^2=0$]