How many triples satisfy $ab + bc + ca = 2 + abc $ 
$a^2 + b^2 + c^2 - \frac{a^3 + b^3 + c^3 - 3abc}{a+b+c} = 2 + abc$
  How many triples $(a,b,c)$ satisfies the statement? Here $a,b,c > 1$.  

It is easy to simplify the statement to
$$ab + bc + ca = 2 + abc.$$
But now how to proceed I don't know. I think this is somehow related to stars and bars theorem but I don't know how to convert this to that problem. Any hint will be helpful.  
Source this is a problem from BdMO 2015 Dhaka regional. 
 A: If $a$, $b$, and $c$ are required to only be positive integers and some of them is $1$, then we have a unique solution $(a,b,c)=(1,1,1)$.  For solutions with $a,b,c>1$, note that
$$(a-1)(b-1)(c-1)=abc-bc-ca-ab+a+b+c-1=a+b+c-3\,.$$
Set $x:=a-1$, $y:=b-1$, and $z:=c-1$. Therefore,
$$xyz=x+y+z\,.$$
Without loss of generality, suppose that $x\leq y\leq z$ (noting that they are positive integers).  Now, we have
$$y+z=x(yz-1)\geq yz-1\,,\text{or }(y-1)(z-1)\leq 2\,.$$
If $(y-1)(z-1)=0$, then $y=1$, making $x=1$ as well, so $$z=xyz=x+y+z=2+z\,$$ which is absurd.  If $(y-1)(z-1)=1$, then $y=2$ and $z=2$, so $$4x=xyz=x+y+z=x+4\,,$$ leading to $x\notin\mathbb{Z}$, which is a contradiction.  Thus, $(y-1)(z-1)=2$, and so $y=2$ and $z=3$.  Ergo, $$6x=xyz=x+y+z=x+5\,,\text{ or }x=1\,.$$
Consequently, $(a,b,c)=(2,3,4)$ is the only solution, up to permutation.
A: I'm also a Bangladeshi, reading in class IX in DRMC. Probably the problem was:

How many solution triads $(a,b,c)$ are there for the equation $a^2 + b^2 + c^2 - \frac{a^3 + b^3 + c^3 - 3abc}{a+b+c} = 2 + abc?$ Where $(a,b,c)$ are positive integers and $a,b,c>1$. 

After trying a lot, I've got a simple solution. Suppose that, $2≤a≤b≤c$.
Note that,
$$ab+bc+ca-2<3bc$$
i.e.
    $$abc<3bc$$      So,  $a=2$
Set the value of $a$. Therefore,
$$2b+2c = 2+bc$$
Now,     $2b+2c=2+bc>bc$
Hence,      $(b-2)(c-2)<4$
Solve the equation. Finally, $(a,b,c)=(2,3,4)$ is the only solution.
