505. Solve the system of equations:

$$
\begin{gathered}
x+y+z=9 \\
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 \\
x y+x z+y z=27
\end{gathered}
$$

Problem from "Mathesis".
505. The second of the given equations can be represented as:

$$
\frac{y z + x z + x y}{x y z} = 1
$$

Substitution from the third equation gives \( x y z = 27 \).

Now we have:

$$
\begin{gathered}
x + y + z = 9 \\
x y + x z + y z = 27 \\
x y z = 27
\end{gathered}
$$

Therefore, \( x, y \) and \( z \) are the roots of the cubic equation

$$
u^{3} - 9 u^{2} + 27 u - 27 = 0
$$

or \((u-3)^{3} = 0\), from which \( u = 3 \), and thus,

$$
x = y = z = 3
$$