2012/025) factor a^2(b-c)^3+b^2(c-a)^3+c^2(a-b)^3.

This is cyclic and we can keep one unchanged and collect the a terms and constants together
= a^2(b-c)^3 + b^2(c^3-3ac^2 + 3a^2c – a^3) + c^2(a^3-3a^2b + 3a b^2 – b^3)
= a^2(b-c)^3 + a^3(c^2-b^2 ) + 3a^2(b^2c – bc^2) + b^2c^3- c^2b^3
= a^3(c^2-b^2) + a^2(b^3 – c^3) + b^2c^2(c-b)
= (c-b)(a^3(c+b) - a^2(b^2 + c^2 + bc) + b^2c^2)

Now
(a^3(c+b) - a^2(b^2 + c^2 + bc) + b^2c^2) is cubic in a but we can make it quadratic in c by decreasing on order of c (easier to factor)
= c^2(b^2-a^2) + c( a^3 – a^2b) + a^3b - a^2b^2
= c^2(b-a)(b+a) + ca^2(a-b) + a^2b(a-b)
= (b-a)(c^2(b+a) + ca^2 – a^2b)

Now we need to factor
(c^2(b+a) - ca^2 – a^2b)
= c^2b + c^2 a – ca^2 – a^2 b
= b(c^2-a^2) + ca(c-a)
= b(c-a) (c+ a) + ca(c-a)
= (c-a) (ab+ bc + ca)

So we get the complete factorization as
(c-a)(c-b)(b-a)(ab+bc + ca)
= (c-a)(b-c)(a-b)(ab + bc +ca) multiplying 2nd and 3rd term by 1 to get in cyclic form