Find all positive integer triples $(x, y, z)$ such that

$$
x y z + x y + y z + z x + x + y + z = 1977.
$$
Notice that by adding 1 to both sides of the equation, the left side can be factored as follows:

$$
\begin{gathered}
x y z + x y + y z + x z + x + y + z + 1 = x(y z + y + z + 1) + (y z + y + z + 1) \\
= (x + 1)(y z + y + z + 1) = (x + 1)(y + 1)(z + 1)
\end{gathered}
$$

Thus, the original equation is equivalent to the following:

$$
(x + 1)(y + 1)(z + 1) = 1978
$$

The values of \(x, y,\) and \(z\) are positive integers, so \(x + 1, y + 1,\) and \(z + 1\) are natural numbers greater than 1. Therefore, we can find their values by decomposing 1978 into the product of three natural numbers greater than 1. Since the prime factorization of 1978 is \(2 \cdot 23 \cdot 43\), there is only one such decomposition. Thus, the values of \(x, y,\) and \(z\) are 1, 22, and 42 in some order. This results in six solutions, where the values of \(x, y,\) and \(z\) are respectively \(1, 22, 42; 1, 42, 22; 22, 1, 42; 22, 42, 1; 42, 1, 22; 42, 22, 1\).