Solve the following system of equations:

$$
\begin{gathered}
x+y+z=1 \\
(y-z)^{2}+(x-z)^{2}+(x-y)^{2}=68 \\
x(y-z)^{2}+y(z-x)^{2}+z(x-y)^{2}=16
\end{gathered}
$$
Let's square (2), rearrange the equation, and divide by 2:

$$
\left(x^{2}+y^{2}+z^{2}-x y-x z-y z\right)=34
$$

Square (1):

$$
x^{2}+y^{2}+z^{2}+2 x y+2 x z+2 y z=1
$$

If we now subtract (2') from (1'), and divide by 3, we get

$$
x y+x z+y z=-11
$$

If we also perform the indicated operations in (3), after rearranging we get

$$
(x y+x z+y z)(x+y+z)-9 x y z=16
$$

In the last line, based on (4) and (1), we know that the minuend is -11, substituting this and rearranging, we get the equation

$$
x y z=-3
$$

If $x, y, z$ are solutions to the system of equations in the problem, then for these three numbers, the following system of equations is also satisfied:

$$
\begin{gathered}
x+y+z=1 \\
x y+y z+x z=-11 \\
x y z=-3
\end{gathered}
$$

Based on the relationships between roots and coefficients, we can now write a cubic equation whose roots are $x$, $y$, and $z$, since

$$
(u-x)(u-y)(u-z)=u^{3}-(x+y+z) u^{2}+(x y+y z+z x) u-x y z
$$

Substituting the values of the coefficients, we get the equation

$$
u^{3}-u^{2}-11 u+3=0
$$

If $x, y$, and $z$ are solutions to the system of equations (1), (2), (3), then these three numbers, in some order, are also roots of equation (6).

Although there is a general method for solving cubic equations, here we can reach our goal more quickly by using the known fact that the rational roots of a polynomial with integer coefficients can be obtained in a finite number of steps: their numerators are divisors of the constant term of the polynomial, and their denominators are divisors of the coefficient of the highest degree term. In our case, the coefficient of $u^{2}$ is 1, so the rational roots of (6) are integers, and they are from the set $\{1, -1, 3, -3\}$.

Substituting -3, it is indeed a root, so the factor $(u+3)$ can be factored out:

$$
u^{3}-u^{2}-11 u+3=(u+3)\left(u^{2}-4 u+1\right)
$$

The second factor is quadratic, with roots $2+\sqrt{3}$ and $2-\sqrt{3}$.

We have found that if $x, y, z$ are solutions to the system of equations in the problem, then the values of $x, y, z$ (in some order) are $-3, 2+\sqrt{3}$, and $2-\sqrt{3}$.

Checking, for example, the value triplet $x=-3, y=2+\sqrt{3}, z=2-\sqrt{3}$, this is indeed a solution to the system of equations. Since the system of equations is symmetric in $x, y, z$, any permutation of these three values is a solution, so there are a total of $3!=6$ solutions. Summarizing the solutions to the system of equations in a table:

| $x$ | -3 | -3 | $2+\sqrt{3}$ | $2+\sqrt{3}$ | $2-\sqrt{3}$ | $2-\sqrt{3}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $y$ | $2+\sqrt{3}$ | $2-\sqrt{3}$ | $2-\sqrt{3}$ | -3 | -3 | $2+\sqrt{3}$ |
| $z$ | $2-\sqrt{3}$ | $2+\sqrt{3}$ | -3 | $2-\sqrt{3}$ | $2+\sqrt{3}$ | -3 |