Find all positive integers $a$ and $b$ such that $(1 + a)(8 + b)(a + b) = 27ab$. Here's the problem I'm having difficulties with:

Find all positive integers $a$ and $b$ such that $$(1 + a)(8 + b)(a + b) = 27ab\,.$$

Does anyone have an idea how to do this? Any detailed solution is welcome! :) 

This is not a complete solution, but it points the way to one, and its simplicity, I think, makes it worth mentioning.
By expanding out the product and rearranging the results, we obtain the equivalent equation to solve,
$${8+8a\over b}+{8+b\over a}=18-a-b$$
Since the left hand side is positive, the right hand side limits the possibilities for $a$ and $b$ to a small enough set for brute force to take over. 
