<s>[INST] Natural numbers $a,b,c$ satisfaying $abc=2(a+b+c)$ How can one find all natural numbers such that: 
$a≤b≤c$
$$abc=2(a+b+c)$$
I tried this :
$abc-2c=2a+2b$ so $c=\frac{2(a+b)}{ab-2}$
 [/INST] The only possible cases are - $(1,3,8) , (1,4,5) , (2,2,4)$
How? I looked at it this way-
$a=1 \implies c=\frac{2b+2}{b-2} $ Check out for $b=3,4$ at $b=4$, $c=5$ ,so you know you have to stop.
$a=2 \implies c=\frac{b+2}{b-1} $ Check out for $b=2,3$ at $b=3$, $c<3$ ,so you know that's enough.
$a=3 \implies c=\frac{2b+6}{3b-2} $ Check out for $b=3$ $c<3$, That's done.
A very tiresome way to do , but it works.
Just a means of simplification, put $a=b=c$ as the boundary condition. You get $a=\sqrt{6}$ Which means you can't go beyond $a=2$ 
Also, in your equation, put $b=c$ to get the boundary condition for $b$ . you get the boundary to be $\frac{2+\sqrt{4+2a}}{a}$ for each $a$
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