Solve the following system of equations:

$$
\begin{gathered}
x^{4}-x^{2} y^{2}-x^{2} z^{2}+y^{2} z^{2}=0 \\
x^{3}-y^{3}-z^{3}=0 \\
x+y+z=1
\end{gathered}
$$
Solve the following system of equations:

$$
\begin{gathered}
x^{4}-x^{2} y^{2}-x^{2} z^{2}+y^{2} z^{2}=0 \\
x^{3}-y^{3}-z^{3}=0 \\
x+y+z=1
\end{gathered}
$$

Transform the left-hand side of the first equation into a product:

$$
\begin{aligned}
x^{4}-x^{2} y^{2}-x^{2} z^{2}+y^{2} z^{2} & =0 \\
\left(x^{2}-y^{2}\right)\left(x^{2}-z^{2}\right) & =0 \\
(x+y)(x-y)(x+z)(x-z) & =0
\end{aligned}
$$

The value of this product is zero if any of the factors is zero. This represents four cases. However, if we notice that the roles of $y$ and $z$ are symmetric (interchangeable) in all three equations, we only need to examine two cases and extend the results to the other two. Let's look at these two cases.

1. $x+y=0$, or $x=-y$. In this case, from equation (3), $z=1$ follows. Using the fact that $x^{3}=-y^{3}$, from equation (2), $y=\sqrt[3]{\frac{1}{2}}$ follows. From the condition, $x=\sqrt[3]{\frac{1}{2}}$ follows.
2. $x-y=0$, or $x=y$. Using the fact that $x^{3}=y^{3}$, from equation (2), $z=0$ follows. Using this, since $x=y$, from equation (3), $x=y=\frac{1}{2}$ follows.

Thus, we have found two solutions that are valid. (We can refer to either verification or equivalent transformations.) Applying the fact that $y$ can be interchanged with $z$, we list all solutions in the following table.

| $x$ |  |  |  |
| :---: | :---: | :---: | :---: |
| $x$ | $z$ |  |  |
| 1 | $\frac{1}{2}$ | $\frac{1}{2}$ | 0 |
| 2 | $\frac{1}{2}$ | 0 | $\frac{1}{2}$ |
| 3 | $\sqrt[3]{\frac{1}{2}}$ | $\sqrt[3]{-\frac{1}{2}}$ | 1 |
| 4 | $\sqrt[3]{\frac{1}{2}}$ | 1 | $\sqrt[3]{-\frac{1}{2}}$ |