Question

(x+yw+zw^2)^2+(xw+yw^2+z)^2+(xw^2+y+zw)^2=0,
prove it if w is a non real cube root of unity.​

Answer 1

Question

$\omega$ is the imaginary solution to the equation of $x^{3}=1$. Prove that $(x+y\omega +z\omega ^{2})^{2}+(x\omega +y\omega ^{2}+z)^{2}+(x\omega ^{2}+y+z\omega )^{2}=0$.

Hint

It is a polynomial identity.

$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(xy+yz+zx)$

The key to the problem is writing $Y,Z$ in terms of $X$.

Solution

Let each term be $X=x+y\omega +z\omega ^{2}$, $Y=x\omega +y\omega ^{2}+z$, and $Z=x\omega ^{2}+y+z\omega$.

We observe that $Y=\omega X$ and $Z=\omega Y$.

$\implies Y=\omega X, Z=\omega ^{2}X$

$\implies X+Y+Z=(\omega ^{2}+\omega +1)X=0$.

By identity, $X^{2}+Y^{2}+Z^{2}=-2(XY+YZ+ZX)$, so $X^{2}+Y^{2}+Z^{2}=-2(\omega X^{2}+\omega ^{3}X^{2}+\omega ^{2}X^{2})$.

$\implies X^{2}+Y^{2}+Z^{2}=-2(\omega ^{3}+\omega ^{2}+\omega )X^{2}$

$\implies X^{2}+Y^{2}+Z^{2}=-2(\omega ^{2}+\omega +1)X^{2}$

$\implies X^{2}+Y^{2}+Z^{2}=0$

$\therefore (x+y\omega +z\omega ^{2})^{2}+(x\omega +y\omega ^{2}+z)^{2}+(x\omega ^{2}+y+z\omega )^{2}=0$ $\blacksquare$