Prove that $a=2,b=2,c=2$ for a system of three equations.

Question

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 $a=b=c=2$.

Answer 1

This is ugly, but you can rewrite the system as.

$$ b=-a^3+3a+4\\ c=-2b^3 + 6b+6 \\ a=-3c^3+9c+8$$

Now, you can substitute the first into the second and that result into the third.

This will show one of the roots, $a = 2$ immediately and another useless one.

You can then sub back in the other two and derive the other two roots.

Answer 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