Solve the equation: $\displaystyle a^2b^2+b^2c^2+c^2a^2+a^2+b^2+c^2+4(a+b+c)+12=6abc+ 4(ab+bc+ca)$
So if $\displaystyle a=b=c$ the equation becomes:
$\displaystyle 3a^4+3a^2+12a+12=6a^3+12a^2\Leftrightarrow a^4-2a^3-3a^2+4a+4=0\Leftrightarrow$ $\displaystyle (-2 + a)^2 (1 + a)^2=0$
So $\displaystyle a=b=c=2$ and $\displaystyle a=b=c=-1$ are solutions.
Substract the second equation from the first:
$\displaystyle (b-c)(a+1)=0$
1. If $\displaystyle b=c$, replace c with b in the first and the third equation:
$\displaystyle \left\{\begin{array}{ll}ab-b=2\\b^2-a=2\end{array}\right.$
From the second equation $\displaystyle a=b^2-2$
Replace a in the first equation: $\displaystyle b^3-3b-2=0\Leftrightarrow (b+1)^2(b-2)=0$
If $\displaystyle b=-1\Rightarrow a=-1, \ c=-1$
If $\displaystyle b=2\Rightarrow a=2, \ c=2$
2. If $\displaystyle a=-1$, replace a in the first and the third equation:
$\displaystyle \left\{\begin{array}{ll}b+c=-2\\bc=1\end{array}\right.\Rightarrow b=c=-1$