of the equation![\\[ (a+b+3)^2 + 2ab = 3ab(a+2)(b+2)\\]](https://latex.artofproblemsolving.com/d/e/1/de172272f7768ba93178467455664a4fd8e6af45.png)
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The idea is that the RHS is a 4 degree polynomial in two variables and the LHS is a 2 degree polynomial in the same two variables.Now, you have the intuition that the 4th-degree polynomial will surpass the LHS after some time. We, therefore, aim to bound the a and b assuming the equality holds.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]So, we try to find when RHS $latex \geq $ LHS. Rather we will find the condition when $latex (a-b)^2 \leq (ab-1)(ab + 2a + 2b + 3) $.Assume $latex a \geq b $.We will find when $latex (a-b) \leq (ab-1)$ and $latex (a-b) \leq (ab + 2a + 2b + 3)$.
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" _i="3" _address="0.1.0.2.3" hover_enabled="0"] $latex (a-b) \leq (ab-1) $ - For this to hold.$latex (a^2-1)(b^2 -1) \geq 0$.$latex (a-b) \leq (ab + 2a + 2b + 3)$ - For this to hold$latex (a+3)(b+1) \geq 0$.
[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" _i="4" _address="0.1.0.2.4" hover_enabled="0"]Hence, try to understand that we have essentially bounded the solutions.Observe this implies that to solutions to have essentially.$latex -3 \leq a,b \leq 1 $.Hence show that by computation that :The solution set is 
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