Let 1 , ω , ω^2 be three cube roots of unity. If x = a + b , y = aω...

Given $x = a + b$, $y = a\omega + b\omega^2$, and $z = a\omega^2 + b\omega$.

Squaring each term, we get:

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

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

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

Using the properties of cube roots of unity, $\omega^3 = 1$ and $\omega^4 = \omega$. Substituting these into the equations:

$y^2 = a^2\omega^2 + b^2\omega + 2ab$

$z^2 = a^2\omega + b^2\omega^2 + 2ab$

Adding $x^2$, $y^2$, and $z^2$:

$x^2 + y^2 + z^2 = a^2(1 + \omega + \omega^2) + b^2(1 + \omega + \omega^2) + 2ab(1 + 1 + 1)$

Since $1 + \omega + \omega^2 = 0$, the expression simplifies to:

$x^2 + y^2 + z^2 = a^2(0) + b^2(0) + 6ab = 6ab$