
3. Find all non-zero real numbers $x, y, z$ which satisfy the system of equations:

$$
\begin{aligned}
\left(x^{2}+x y+y^{2}\right)\left(y^{2}+y z+z^{2}\right)\left(z^{2}+z x+x^{2}\right) & =x y z \\
\left(x^{4}+x^{2} y^{2}+y^{4}\right)\left(y^{4}+y^{2} z^{2}+z^{4}\right)\left(z^{4}+z^{2} x^{2}+x^{4}\right) & =x^{3} y^{3} z^{3}
\end{aligned}
$$


Solution: Since $x y z \neq 0$, We can divide the second relation by the first. Observe that

$$
x^{4}+x^{2} y^{2}+y^{4}=\left(x^{2}+x y+y^{2}\right)\left(x^{2}-x y+y^{2}\right)
$$

holds for any $x, y$. Thus we get

$$
\left(x^{2}-x y+y^{2}\right)\left(y^{2}-y z+z^{2}\right)\left(z^{2}-z x+x^{2}\right)=x^{2} y^{2} z^{2}
$$

However, for any real numbers $x, y$, we have

$$
x^{2}-x y+y^{2} \geq|x y|
$$

Since $x^{2} y^{2} z^{2}=|x y||y z||z x|$, we get

$$
|x y||y z||z x|=\left(x^{2}-x y+y^{2}\right)\left(y^{2}-y z+z^{2}\right)\left(z^{2}-z x+x^{2}\right) \geq|x y||y z||z x|
$$

This is possible only if

$$
x^{2}-x y+y^{2}=|x y|, \quad y^{2}-y z+z^{2}=|y z|, \quad z^{2}-z x+x^{2}=|z x|
$$

hold simultaneously. However $|x y|= \pm x y$. If $x^{2}-x y+y^{2}=-x y$, then $x^{2}+y^{2}=0$ giving $x=y=0$. Since we are looking for nonzero $x, y, z$, we conclude that $x^{2}-x y+y^{2}=x y$ which is same as $x=y$. Using the other two relations, we also get $y=z$ and $z=x$. The first equation now gives $27 x^{6}=x^{3}$. This gives $x^{3}=1 / 27$ (since $x \neq 0$ ), or $x=1 / 3$. We thus have $x=y=z=1 / 3$. These also satisfy the second relation, as may be verified.
