5. Indeterminate Equation
$$
\begin{array}{l}
\frac{1}{2}(x+y)(y+z)(z+x)+(x+y+z)^{3} \\
=1-x y z
\end{array}
$$

The number of integer solutions to the equation is ( ).
(A) 2
(B) 4
(C) 5
(D) 6
5.D.

Let $x+y=u, y+z=v, z+x=w$.
Then the equation transforms to
$$
\begin{array}{l}
4 \text { uvw }+(u+v+w)^{3} \\
=8-(u+v-w)(u-v+w) . \\
\quad(-u+v+w) .
\end{array}
$$

Simplifying, we get
$$
4\left(u^{2} v+v^{2} w+w^{2} u+u v^{2}+v w^{2}+u u^{2}\right)+8 u w=8 \text {, }
$$

which is
$$
u^{2} v+v^{2} w+w^{2} u+w w^{2}+w w^{2}+w u^{2}+2 w w=2 \text {. }
$$

Factoring the left side, we get
$$
\begin{array}{l}
(u+v)(v+w)(w+u)=2 . \\
\text { Thus, }(u+v, v+w, w+u) \\
=(1,1,2),(-1,-1,2),(1,-2,-1)
\end{array}
$$

and symmetric cases.
Solving each case, we get
$$
(u, v, w)=(1,0,1),(1,-2,1),(1,0,-2) \text {. }
$$

Thus, $(x, y, z)=(1,0,0),(2,-1,-1)$.
In summary, considering symmetry, the original equation has 6 integer solutions.