3. Find all integer solutions $a, b, c$ that satisfy $\frac{1}{2}(a+b)(b+c)(c+a)+(a+b+c)^{3}=1-a b c$, and provide necessary proof.

Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
3. Solution: Let $s=a+b+c, p(x)=(x-a)(x-b)(x-c)=x^{3}-s x^{2}+(a b+b c+c a) x-a b c$.

Then $(a+b)(b+c)(c+a)=p(s)=(a b+b c+c a) s-a b c$.
Thus, the given equation can be rearranged as $(a b+b c+c a) s-a b c=-2 s^{3}+2-2 a b c$,
which simplifies to $p(-s)+2=0$.
This indicates that $(2 a+b+c)(a+2 b+c)(a+b+2 c)=2$.
It is easy to see that one of the factors on the left side is 2, and the other two factors are 1,1 (or -1,-1), or one factor is -2, and the other two factors are 1,-1 (or -1,1).
Solving these, we get $a=1, b=0, c=0 ; a=2, b=-1, c=-1$.
When one of the factors is -2, the corresponding two equations can lead to $4(a+b+c)=-2$, which clearly has no integer solutions $(a, b, c)$.

Since the equation is symmetric in $a, b, c$, the complete set of possible values for $(a, b, c)$ is $(1,0,0),(0,1,0)$, $(0,0,1),(2,-1,-1),(-1,2,-1),(-1,-1,2)$.