For the variables $a=x, b=x+y, c=x+y+z$, (1) represents the system of equations

$$
a+b+c=10, \quad a b+b c+c a=27, \quad a b c=18
$$

According to (2), the numbers $a, b, c$ are the roots of the equation

$$
z^{3}-10 z^{2}+27 z-18=0
$$

in some order, since on the left side of (3) due to (2), $(z-a)(z-b)(z-c)$ stands. Since the value on the left side of (3) is 0 for $z=1$, $z-1$ can be factored out from it:

$$
z^{3}-10 z^{2}+27 z-18=(z-1)\left(z^{2}-9 z+18\right)
$$

The roots of the remaining quadratic factor are 3 and 6. Therefore, the roots of (2) are:

$$
\begin{array}{rrrrrrr} 
& a=1, & 1, & 3, & 3, & 6, & 6 ; \\
& b=3, & 6, & 1, & 6, & 1, & 3 ; \\
\text { and the corresponding roots of (1) are: } \quad r=6, & 3, & 6, & 1, & 3, & 1 \text {; } \\
& & & & & & \\
& x=a=1, & 1, & 3, & 3, & 6, & 6 ; \\
y=b-a=2, & +5, & -2, & 3, & -5, & -3 ; \\
z=c-b=3, & -3, & 5, & -5, & 2, & -2 \text {; }
\end{array}
$$
Solve the following system of equations:

$$
\begin{aligned}
3 x+2 y+z & =10 \\
3 x^{2}+4 x y+2 x z+y^{2}+y z & =27 \\
x^{3}+2 x^{2} y+x^{2} z+x y^{2}+x y z & =18
\end{aligned}
$$