This was a bit hard to crack. I'll show you my thought chain. By the nature of the equation you see there are going to be independent a, b, c terms in the equation when expanded. This hints to a final factor containing constant term (±1) multiplied by another which contains independent terms (± a ± b ± c). After that I couldn't make any conclusions to decide the factors, and hence expanded the equation, which gives,
a²b²c + b²c²a + c²a²b - a²b - a²c - b²a - b²c - c²a - c²b + a + b + c - 4abc
Taking out ab, bc and ca as factors, and by cyclic symmetry in the equation,
ab(abc - a - b) + bc(abc - b - c) + ca(abc - c - a) + (a + b + c - 4abc)
Now if the factors (abc - a - b) , (abc - b - c) and (abc - c - a) could all be made of the form (abc - a - b - c) you can factor it out from all the terms. See if you can do this.
I'm not aware of any specific articles on factorization. Just play around with enough problems to give you an hint of what you can do in various situations.
PS: I'm curious to see if there are more elegant solutions to this from someone, than expanding out the whole equation.
Edit: Google gave these articles,
http://tutorial.math.lamar.edu/Classes/Alg/Factoring.aspx
http://www.qc.edu.hk/math/Resource/AL/Cyclic%20and%20symmetric%20polynomials.pdf