736. Find all representations of the number 1 in the form of the sum

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

where $x, y, z$ are natural numbers.
$\triangle$ Given:

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

Let $x \leqslant y \leqslant z$. Then among the three products $x y, x z$, and $y z$, the largest is $y z$. Using this, we obtain the inequality:

$$
x y z=x y+x z+y z \leqslant y z+y z+y z=3 y z
$$

Thus, $x y z \leqslant 3 y z$, from which $x \leqslant 3$. Since $x>1$, $x$ can only take the values 2 and 3. Let's consider both cases.

1) Suppose $x=2$. Then the equation becomes

$$
2 y z=2 y+2 z+y z, \quad y z=2 y+2 z .
$$

The last equation can be solved using inequalities (try to do this yourself), but there is another way. Express $z$ in terms of $y$:

$$
y z-2 z=2 y, \quad z=\frac{2 y}{y-2}
$$

Find all values of $y$ for which the last fraction is an integer. As we saw in § 8 (point 8.1.6), we get:

$$
z=\frac{(2 y-4)+4}{y-2}=2+\frac{4}{y-2}
$$

Thus, $y-2$ can take the values 1, 2, and 4.

If $y-2=1$, then $y=3, z=2+4=6$.

If $y-2=2$, then $y=4, z=2+\frac{4}{2}=4$.

If $y-2=4$, then $y=6, z=2+\frac{4}{4}=3$. We temporarily discard this option because $y \leqslant z$ must hold.

2) Suppose $x=3$. Then

$$
3 y z=3 y+3 z+y z, \quad 2 y z=3 y+3 z
$$

Express $z$ in terms of $y$:

$$
2 y z-3 z=3 y, \quad z=\frac{3 y}{2 y-3}
$$

Multiply the last equation by 2 and separate the integer part of the resulting fraction on the right:

$$
2 z=\frac{6 y}{2 y-3}=\frac{(6 y-9)+9}{2 y-3}=3+\frac{9}{2 y-3}
$$

Thus, $2 y-3$ equals 1, 3, or 9.

If $2 y-3=1$, then $y=2$. We discard this option because $x \leqslant y$ must hold.

If $2 y-3=3$, then $y=3, 2 z=3+\frac{9}{3}=6, z=3$.

If $2 y-3=9$, then $y=6, 2 z=3+\frac{9}{9}=4, z=2$. We discard this option as well.

Answer: $1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{1}{2}+\frac{1}{4}+\frac{1}{4}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}$