We are given two equations, $a+b=1-c \tag{1}$ $a^{3}+b^{3}=1-c^{2} \tag{2}$

Rewriting $(2)$ factored as: $(a+b)(a^{2}-ab+b^{2}) = (1-c)(1+c)$

Hence, $(2) \div (1)$ attains, $a^{2}-ab+b^{2} = 1+c \tag{3}$

Adding $(1)$ to $(3)$ and collecting terms on one side we obtain, $a^{2}-ab-a+b^{2}+b-2=0 \tag{4}$

We can play with $(4)$ to obtain the following, which can be used in a quadratic formula: $a^{2} + a(1-b) + (b+2)(b-1) = 0$

Using the quadratic formula, $a = \frac{b-1 \pm \sqrt{(1-b)^{2}-4(b+2)(b-1)}}{2}$

The discriminant shouldn't be negative: $discriminant \ge 0$

Hence, $(1-b)^{2}-4(b+2)(b-1) \ge 0$

Expanding the LHS and simplifying it afterwards, $(b+3)(b-1) \le 0$

Solving this quadratic inequality we find that: $-3 \le b \le 1$ Thus, b has 5 possible solutions,

\tag{5} $a = \frac{b-1 \pm \sqrt{(1-b)^{2}-4(b+2)(b-1)}}{2}$

$\underline{Case\:b=-3:}$

Using $(5)$ we get $a = -2$

Hence, using $(1)$ to get $\; c = 6$

$\underline{Case\:b=-2:}$

Using $(5)$ we get $a = 0,-3$

For $a= 0$ , $\; c=3$

For $a=-3$ , $\; c=6$

$\underline{Case\:b=-1:}$

Using $(5)$ we get $a = -1+\sqrt{3}$

Hence, this solution is rejected as a is not an integer.

$\underline{Case\:b=0:}$

Using $(5)$ we get $a = 1.-2$

For $a= 1$ , $\; c=0$

For $a=-2$ , $\; c=3$

$\underline{Case\:b=1:}$

Using $(5)$ we get $a = 0$

Hence, using $(1)$ to get $\; c = 0$

Finally, we have a total of 6 solutions, $(1,0,0) , (-2,0,3), (0,1,0), (-2,-3,6), (0,-2,3), (-3,-2,6)$

Hence the answer, $1-2+3+1-2-3+6-2+3-3-2+6 = \boxed{6}$

If $a+b=0$ then $(t,-t,1)$ is a family of solutions

So suppose $a+b≠0$ and hence $a^{2}-ab+b^{2}+a+b-2=0$

The equation can be written as $(2a-b+1)^{2}+3(b+1)^{2}=12$

There are two possibilities $(2a-b+1)= _{-}^{+}3$ , $(b+1)= _{-}^{+}1$ and $(2a-b+1)= 0$ , $(b+1)= _{-}^{+}2$

Hence the solution sets are $(0,1,0),(-2,-3,6),(1,0,0),(0,-2,3),(-2,0,3),(-3,-2,6)$