I want to prove that the following equation has no integer solutions $a,b,c$: $$-a^3 - b^3 - c^3 + ab^2 - ac^2 + bc^2 - 2a^2c + 3abc = 0$$ apart from the naive solution $a=b=c=0$.
The context, in case it helps, is the following: I am dealing with the field $\mathbb{F}_p[x] / (x^3-x-1)$ and this equation appeared when I tried to get a "formula" for the inverse of some element on this field. More precisely, if $g(x) = a + bx +cx^2$ is an element of the field and $g(x)' = c + dx + e^2$ is assumed to be the inverse of $g(x)$ I have tried to compute $c,d$ and $e$ as a function of $a,b$ and $c$. The result being: $$d = \frac{a²+2ac-bc-b²+c²}{-a^3 - b^3 - c^3 + ab^2 - ac^2 + bc^2 - 2a^2c + 3abc}, \\ e = \frac{ba-c²}{-a^3 - b^3 - c^3 + ab^2 - ac^2 + bc^2 - 2a^2c + 3abc}, \\ f = \frac{-b²+ca+c²}{-a^3 - b^3 - c^3 + ab^2 - ac^2 + bc^2 - 2a^2c + 3abc}.$$
However, the expression has a denominator that happens to be the above equation. So, as $\mathbb{F}_p[x] / (x^3-x-1) = \mathbb{F}_{p^3}$ we know that every element on this field has to have a multiplicative inverse (except the $0$) and therefore the above equation cannot be $0$.
How can I be more precise in the proof?
For a contradiction assume $a, b, c$ is a solution.
First of all, if $a, b, c$ have any common factor, we can factor it out so without loss generality assume $a, b, c$ have no common factor.
Now, reduce the equation $\mod 2$: \begin{align} -a^3 - b^3 -c^3 + ab^2 - ac^2 + bc^2 -2a^2c + 3abc &\equiv\\ a + b + c + ab + ac + bc + abc &\equiv \\ (1 + a)(1 + b)(1 + c) - 1 &\equiv 0 \mod 2 \end{align}
But this shows, $a, b, c$ are all even which is impossible because $a, b, c$ have no common factor.