Find all non-zero real numbers $ x, y, z$ which satisfy the system of equations:
\[ (x^2 \plus{} xy \plus{} y^2)(y^2 \plus{} yz \plus{} z^2)(z^2 \plus{} zx \plus{} x^2) \equal{} xyz\]
\[ (x^4 \plus{} x^2y^2 \plus{} y^4)(y^4 \plus{} y^2z^2 \plus{} z^4)(z^4 \plus{} z^2x^2 \plus{} x^4) \equal{} x^3y^3z^3\]
To solve the given system of equations, we need to find all non-zero real numbers \( x, y, z \) that satisfy:

\[
(x^2 + xy + y^2)(y^2 + yz + z^2)(z^2 + zx + x^2) = xyz
\]

\[
(x^4 + x^2y^2 + y^4)(y^4 + y^2z^2 + z^4)(z^4 + z^2x^2 + x^4) = x^3y^3z^3
\]

1. **Analyzing the first equation:**

   \[
   (x^2 + xy + y^2)(y^2 + yz + z^2)(z^2 + zx + x^2) = xyz
   \]

   We can use the AM-GM inequality on each term inside the product:

   \[
   x^2 + xy + y^2 \geq 3\sqrt[3]{x^2 \cdot xy \cdot y^2} = 3xy
   \]

   Similarly,

   \[
   y^2 + yz + z^2 \geq 3yz
   \]

   \[
   z^2 + zx + x^2 \geq 3zx
   \]

   Multiplying these inequalities, we get:

   \[
   (x^2 + xy + y^2)(y^2 + yz + z^2)(z^2 + zx + x^2) \geq 27x^2y^2z^2
   \]

   Given that the left-hand side equals \( xyz \), we have:

   \[
   xyz \geq 27x^2y^2z^2
   \]

   Simplifying, we get:

   \[
   1 \geq 27xyz
   \]

   \[
   xyz \leq \frac{1}{27}
   \]

2. **Analyzing the second equation:**

   \[
   (x^4 + x^2y^2 + y^4)(y^4 + y^2z^2 + z^4)(z^4 + z^2x^2 + x^4) = x^3y^3z^3
   \]

   Using the AM-GM inequality again:

   \[
   x^4 + x^2y^2 + y^4 \geq 3\sqrt[3]{x^4 \cdot x^2y^2 \cdot y^4} = 3x^2y^2
   \]

   Similarly,

   \[
   y^4 + y^2z^2 + z^4 \geq 3y^2z^2
   \]

   \[
   z^4 + z^2x^2 + x^4 \geq 3z^2x^2
   \]

   Multiplying these inequalities, we get:

   \[
   (x^4 + x^2y^2 + y^4)(y^4 + y^2z^2 + z^4)(z^4 + z^2x^2 + x^4) \geq 27x^4y^4z^4
   \]

   Given that the left-hand side equals \( x^3y^3z^3 \), we have:

   \[
   x^3y^3z^3 \geq 27x^4y^4z^4
   \]

   Simplifying, we get:

   \[
   1 \geq 27xyz
   \]

   \[
   xyz \leq \frac{1}{27}
   \]

3. **Combining the results:**

   From both equations, we have:

   \[
   xyz \leq \frac{1}{27}
   \]

   and

   \[
   xyz \geq \frac{1}{27}
   \]

   Therefore,

   \[
   xyz = \frac{1}{27}
   \]

4. **Checking for equality:**

   Equality in the AM-GM inequality holds when all terms are equal. Thus, we have:

   \[
   x^2 = xy = y^2 \implies x = y
   \]

   Similarly,

   \[
   y^2 = yz = z^2 \implies y = z
   \]

   Therefore, \( x = y = z \). Substituting \( x = y = z \) into \( xyz = \frac{1}{27} \), we get:

   \[
   x^3 = \frac{1}{27} \implies x = \frac{1}{3}
   \]

   Hence, the solution is:

   \[
   x = y = z = \frac{1}{3}
   \]

The final answer is \(\boxed{\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)}\).