3B. Solve the system

$$
\left\{\begin{array}{l}
x(y+z)=35 \\
y(z+x)=32 \\
z(x+y)=27
\end{array}\right.
$$
Solution. If we develop the system to the form

$$
\left\{\begin{array}{l}
x y + x z = 35 \\
y z + y x = 32 \\
z x + z y = 27
\end{array}\right.
$$

it is obvious that we can introduce substitutions $a = xy, b = xz, c = yz$, so the system takes the form

$$
\left\{\begin{array}{l}
a + b = 35 \\
c + a = 32 \\
b + c = 27
\end{array}\right.
$$

By adding and subtracting the equations, we get:

$$
\begin{aligned}
& (1) + (2) - (3): 2a = 40, \\
& (2) + (3) - (1): 2c = 12 \text{ and } \\
& (1) + (3) - (2): 2b = 30.
\end{aligned}
$$

From here, the corresponding system becomes

$$
\left\{\begin{array}{l}
a = 20 \\
b = 15 \\
c = 12
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
x y = 20 \\
x z = 15 \\
y z = 12
\end{array}\right.
$$

and at the same time $x, y, z \neq 0$. A system equivalent to the given one is

$$
\left\{\begin{array}{l}
y = \frac{20}{x} \\
z = \frac{15}{x} \\
y z = 12
\end{array}\right.
$$

Substituting the first and second equations into the third, we get $x^2 = 25, x = \pm 5$.

Then the solutions to the original system are

$$
(x, y, z) \in \{(5, 4, 3), (-5, -4, -3)\}
$$