Solve the following system of equations:

$$
\left\{\begin{array}{l}
x \cdot(y+z) \cdot(x+y+z)=1170 \\
y \cdot(z+x) \cdot(x+y+z)=1008 \\
z \cdot(x+y) \cdot(x+y+z)=1458
\end{array}\right.
$$
Let $x+y+z=u$, and rewrite the equations in the following form:

$$
\begin{aligned}
& x y u + x z u = 1170 \\
& x y u + y z u = 1008 \\
& x z u + y z u = 1458
\end{aligned}
$$

This is a system of linear equations in the unknowns $x y u, x z u, y z u$; the sum of the three unknowns is $(1170 + 1008 + 1458) / 2 = 1818$, so

$$
y z u = 1818 - 1170 = 648, \quad x y u = 360, \quad x z u = 810
$$

According to the original equations, none of the factors here are 0, so dividing the first product by the latter two gives the ratio of the pairs of the original unknowns:

$$
\frac{z}{x} = \frac{648}{360} = \frac{9}{5}, \quad \frac{y}{x} = \frac{4}{5}, \quad z = \frac{9 x}{5}, \quad y = \frac{4 x}{5}
$$

Finally, substituting these expressions into, for example, the first equation

$$
x \cdot \frac{13 x}{5} \cdot \frac{18 x}{5} = 1170, \quad x^{3} = 125
$$

the only real solution to this is $x = 5$, and thus $y = 4, z = 9$.

We always transformed to equivalent equations, so the solution satisfies (1).

Remark. Denoting the constants on the right-hand side by $a, b, c$ and their sum by $2s$, similarly

$$
\begin{gathered}
x y u = s - c, \quad x z u = s - b, \quad y z u = s - a \\
y = \frac{s - a}{s - b} \cdot x, \quad z = \frac{s - a}{s - c} \cdot x \\
\frac{(s - a)^{2}}{(s - b)(s - c)}\left(\frac{1}{s - a} + \frac{1}{s - b} + \frac{1}{s - c}\right) x^{3} = 1 \\
x = \frac{1}{s - a} \sqrt[3]{\frac{(s - a)^{2}(s - b)^{2}(s - c)^{2}}{a b + a c + b c - s^{2}}}
\end{gathered}
$$

provided that $a, b, c$ are not equal to the sum of the other two.