Expand the expression:
a^2b^3-a^3b^2+b^2c^3-b^3c^2+c^2a^3-c^3a^2.
Careful inspection shows that the expression is symmetrical in all three variables. Also, if the expression is equal to zero, putting a=b makes the expression zero, so does b=c and c=a. Therefore (a-b), (b-c) and (c-a) must be factors. Therefore part of the factorisation is (a-b)(b-c)(c-a). But when this is expanded (see below) the sum total of powers in the expansion is 3, but we require a sum equal to 5. For example, we have ab^2 but we need a^2b^3, so we need to multiply by ab. We still need symmetry so we try ab+bc+ac as another factor (see below).
(a-b)(b-c)(c-a)(ab+bc+ca)=(ab^2-a^2b+bc^2-b^2c+ca^2-c^2a)(ab+bc+ca)=
a^2b^3-a^3b^2+ab^2c^2-ab^3c+a^3bc-a^2bc^2+
ab^3c-a^2b^2c+b^2c^3-b^3c^2+a^2bc^2-abc^3+
a^2b^2c-a^3bc+abc^3-ab^2c^2+a^3c^2-a^2c^3=
a^2b^3-a^3b^2+b^2c^3-b^3c^2+a^3c^2-a^2c^3.
This matches the original expression, because all the other terms cancel out.