Solve the following system of equations:

$$
\begin{aligned}
x+y+z & =2 \\
x^{3}+y^{3}+z^{3} & =20 \\
x^{7}+y^{7}+z^{7} & =2060
\end{aligned}
$$
I. solution. We introduce the following notations:

$$
s_{k}=x^{k}+y^{k}+z^{k} \quad(k \geqq 0)
$$

and

$$
a=x y+x z+y z, \quad b=x y z
$$

Our "strategy" is as follows: We know the value of $x+y+z$. We will show that for any natural number $k$, $s_{k}$ can be expressed using this value, as well as $a$ and $b$. Since we know the values of $s_{3}$ and $s_{7}$, we will get two equations for $a$ and $b$. It turns out that this system of two equations with two unknowns is very easy to solve.

First, let's write down the cubic equation whose roots are precisely $x, y$, and $z$:

$$
(u-x)(u-y)(u-z)=0
$$

Expanding and rearranging this, we get the equation

$$
u^{3}=2 u^{2}-a u+b
$$

If $x, y, z$ is a solution triplet of the system of equations in the problem, then any of them will be a root of (4), and indeed, by multiplying the powers of $u$ in (4), we get the equations

$$
u^{k}=2 u^{k-1}-a u^{k-2}+b u^{k-3} \quad(k \geqq 3)
$$

This means that

$$
\begin{aligned}
x^{k} & =2 x^{k-1}-a x^{k-2}+b x^{k-3} \\
y^{k} & =2 y^{k-1}-a y^{k-2}+b y^{k-3}, \\
z^{k} & =2 z^{k-1}-a z^{k-2}+b z^{k-3}, \quad k \geqq 3
\end{aligned}
$$

Adding these three equations, we get the relationship

$$
s_{k}=2 s_{k-1}-a s_{k-2}+b s_{k-3,} \quad k \geqq 3
$$

Thus, we can sequentially calculate the values of $s_{3}, s_{4}, s_{5}, s_{6}, s_{7}$ using $a$ and $b$, since $s_{0}=3, s_{1}=2$ are given, and from this,

$$
s_{2}=x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(x y+x z+y z)=4-2 a
$$

Let's see what (5) gives for $s_{3}$!

$$
s_{3}=2 s_{2}-a s_{1}+b s_{0}=2(4-2 a)-2 a+3 b
$$

According to (2), $s_{3}=20$, substituting this and simplifying, we get the relationship $b=4+2 a$.

Our initial plan thus simplifies significantly: $s_{k}$ can now be expressed using only $a$:

$$
s_{k}=2 s_{k-1}-a s_{k-2}+(4+2 a) s_{k-3}, \quad k \geqq 3
$$

Applying this for $k=4,5,6,7$:

$$
\begin{aligned}
s_{4} & =2 s_{3}-a s_{2}+(4+2 a) s_{1}=40-a(4-2 a)+2(4+2 a)=48+2 a^{2} \\
s_{5} & =2 s_{4}-a s_{3}+(4+2 a) s_{2}=2\left(48+2 a^{2}\right)-20 a+(4+2 a)(4-2 a)=112-20 a \\
s_{6} & =2 s_{5}-a s_{4}+(4+2 a) s_{3}=2(112-20 a)-a\left(48+2 a^{2}\right)+(4+2 a) \cdot 20= \\
& =304-48 a-2 a^{3} . \\
s_{7} & =2 s_{6}-a s_{5}+(4+2 a) s_{4}=2\left(304-48 a-2 a^{3}\right)-a(112-20 a)+ \\
& +(4+2 a)\left(48+2 a^{2}\right)=800-112 a+28 a^{2}
\end{aligned}
$$

Since the value of $s_{7}=2060$ is known, we get a quadratic equation for $a$. After simplification, we get $a^{2}-4 a-45=0$, from which $a_{1}=-5$ or $a_{2}=9$. The latter does not provide real solutions for $x, y, z$ because $0 \leqq s_{2}=x^{2}+y^{2}+z^{2}=4-2 a=-14$ is impossible. Thus, we get only one real solution for $a$ and $b$: $a=-5, \quad b=-6$. The cubic equation is therefore:

$$
u^{3}-2 u^{2}+6=0
$$

and its solutions are 1, -2, and 3. The possible solutions to our system of equations are the six permutations of these three numbers:

$$
\begin{array}{llllll}
x_{1}=3 ; & x_{2}=3 ; & x_{3}=1 ; & x_{4}=1 ; & x_{5}=-2 ; & x_{6}=-2 \\
y_{1}=1 ; & y_{2}=-2 ; & y_{3}=3 ; & y_{4}=-2 ; & y_{5}=3 ; & y_{6}=1 \\
z_{1}=-2 ; & z_{2}=1 ; & z_{3}=-2 ; & z_{4}=3 ; & z_{5}=1 ; & z_{6}=3
\end{array}
$$

It is easy to verify that for these triplets, $s_{1}=2, s_{3}=20, s_{7}=2060$, so they are indeed solutions to the system of equations.

Remark. It is interesting to note that although $s_{7}$ is a seventh-degree polynomial, we were able to express it as a quadratic polynomial in $a$, while $s_{1}$ and $s_{3}$ are only quadratic (products of second-degree terms). We were able to reduce the degree now because we knew the values of $s_{1}$ and $s_{3}$. The same situation occurs with $s_{5}$, where the right-hand side is only linear in $a$. However, the degree does not decrease for $s_{4}$ or $s_{6}$. It is easy to verify that if we continued to express $s_{k}$ in terms of $a$ (without knowing the value of $s_{7}$), for even $k$, we would always get a polynomial of degree $\frac{k}{2}$ in $a$, with the coefficient of $a^{k / 2}$ alternating between +2 and -2, and for odd $k$, the degree would decrease: it would be $\frac{k-3}{2}$ in $a$. Thus, for example, a system of equations where the value of $s_{11}$ is given instead of $s_{7}$ is theoretically solvable, but not always if $s_{10}$ is given, since it would require solving a fourth-degree equation, and there is no general method for solving a fifth-degree equation. (This is true regardless of the specific values of $s_{1}$ and $s_{3}$.)

II. solution. Let $a=x y+y z+z x$ and $b=x y z$ again. This time, we will determine the coefficients of the cubic polynomial

$$
u^{3}-2 u^{2}+a u-b
$$

whose roots are the solutions $x, y, z$ of the system of equations. For this, we use the following identities:

$$
\begin{gathered}
(x+y+z)^{3}-\left(x^{3}+y^{3}+z^{3}\right)=3(x+y)(y+z)(z+x)= \\
=(x+y+z)(x y+y z+z x)-x y z \\
(x+y+z)^{7}+\left(x^{7}+y^{7}+z^{7}\right)= \\
=7(x+y)(y+z)(z+x)\left[\left(x^{2}+y^{2}+z^{2}+x y+y z+z x\right)^{2}+x y z(x+y+z)\right]
\end{gathered}
$$

The left