㚞 Example 2 Find the integer solutions of the indeterminate equation $\frac{1}{2}(x+y)(y+z)(z+x)+(x+y+z)^{3}=1-x y z$.
Let's make the substitution, set $x+y=u, y+z=v, z+x=w$, then the original equation transforms to
$4 u r w+(u+v+w)^{3}=8-(u+v-w)(u-v+w)(-u+v+w)$,
after expanding and combining like terms, we get
$4\left(u^{2} v+v^{2} w+w^{2} u+u v^{2}+v w^{2}+w u^{2}\right)+8 u v w=8$,
which simplifies to $u^{2} v+v^{2} w+w^{2} u+u v^{2}+v w^{2}+w u^{2}+2 u v w=2$.
Factoring the left side of the equation, we get $(u+v)(v+w)(w+u)=2$.
Thus, $(u+v, v+w, w+u)=(1,1,2),(-1,-1,2),(-2,-1,1)$ and their symmetric cases. Solving each, we get $(u, v, w)=(1,0,1),(1,-2,1),(-1,0,2)$, which leads to $(x, y, z)=(1,0,0), (2,-1,-1)$.
In summary, considering symmetry, the integer solutions to the original equation are $(x, y, z)=(1,0,0),(0,1,0)$, $(0,0,1),(2,-1,-1),(-1,2,-1),(-1,-1,2)$, a total of 6 solutions.