Prove that $a=2,b=2,c=2$ for a system of three equations. Let us consider a   system equations :
\begin{gather}  
a^3+b=3a+4\tag{i}\\
2b^3+c=6b+6 \tag{ii}\\
3c^3+a=9c+8\tag{iii}
\end{gather}
I have tried with the following steps :
$6\times(i)+3\times(ii)+2\times$(iii) gives :
$6(a^3+b^3+c^3)+6b+3c+2a=18(a+b+c)+58$. But  I don't know how to solve in this way. By inspection method I have seen $a=b=c=2$.  

Write the equations as
\begin{align*}
(a-2)(a+1)^2 = 2-b\\
2(b-2)(b+1)^2 = 2-c\\
3(c-2)(c+1)^2 = 2-a
\end{align*}
Note that $(a,b,c) = (2,2,2)$ is a valid solution.
$a>2 \Rightarrow b<2\Rightarrow c>2 \Rightarrow a<2. $ A similar logic precludes $a<2$. This forces $a=b=c=2$ which is the unique solution
