For complex $z$, let $N(z)$ denote the norm of $z$, $N(z)=z \overline{z}$ where $\overline{z}$ is the complex conjugate of $z$.
One day, one of my colleague ask me if I can solve the following.
Let $\omega$ be the primitive third root of unity in the field of complex numbers; $\omega^3 =1$, $\omega \neq 1$. For complex numbers $x$, $y$, $z$, let $A = x+y+z$, $B=x+y\omega+z\omega^2$, $C=x+y\omega^2 +z\omega$. Prove that $$ N(A) + N(B) + N(C) = 3(N(x) + N(y)+N(z) $$
Of course he already have a solution by direct calculation involving
$$|A|^2+|B|^2+|C|^2=3(|x|^2+|y|^2+|z|^2)+ 2{\rm Re} (x\overline{y}+x\overline{z}+y\overline{z}+\overline{x}y\omega+\overline{x}z\omega^2+\overline{y\omega}z\omega^2+\overline{x}y\omega^2+\overline{x}z\omega+\overline{y\omega^2}z\omega)$$
The Question Is there any more illuminating approach to this problem?
The problem is equivalent to the following.
Consider the equation $$ t^3 - 3xt^2 + (3x^2-3yz) t + x^3+y^3+z^3+3xyz =0 $$ where $x$, $y$, $z \in \mathbb{C}$. Prove that the sum of the norm of the root of the equation equals to sum of the norm of $x$, $y$, $z$ multiplied by $3$.
Thanks for your attention.