Question

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

$(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)=xyz$ ,

$(x^4+x^2{y}^2+y^4)(y^4+y^2{z}^2+z^4)(z^4+z^2{x}^2+x^4)=x^3{y}^3{z}^3$ .

Answer 1

Since $xyz\ne 0$

We can divide the second relation by first
$x^4+x^2{y}^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2)$ ,

We get,
$(x^2-xy+y^2)(y^2-yz+z^2)(z^2-zx+x^2)=x^2{y}^2{z}^2$.

However, for any real numbers $x,y,$ we have
$x^2-xy+y^2\ge |xy|$.

Since $x^2{y}^2{z}^2=|xy||yz||zx|$, we get
$|xy||yz||zx|=(x^2-xy+y^2)(y^2-yz+z^2)(z^2-zx+x^2)\ge |xy||yz||zx|$.

This is possible only if,
$x^2-xy+y^2=|xy|,y^2-yz+z^2=|yz|,z^2-zx+x^2=|zx|$ hold

simultaneously.

However, $|xy|=\pm xy$ . If $x^2-xy+y^2=-xy$ , then $x^2+y^2=0$ giving $x=y=0$.

Since we are looking for nonzero $x,y,z$ we conclude that $x^2+y^2=xy$ which is same as $x=y$.

Using the other two relations, we get $y=z$ and $z=x$.

The first equation now gives $27x^6=x^3$

$\therefore x^3=1/27$ (since $x\ne 0$ ), or $x=1/3$

$\therefore x=y=z=1/3$.

These also satisfy the second.