Factorization of cyclic polynomial 
Factorize $$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$ 

Since this is a cyclic polynomial, factors are also cyclic 
$$f(a) = a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$f(b) = b(b^2-c^2)+b(c^2-b^2)+c(b^2-b^2) = 0 \Rightarrow a-b$$
is a factor of the given expression. Therefore, other factors are $(b-c)$ and $(c-a)$. The given expression may have a coefficient a constant factor which is nonzero. Let it be $m$.
$$\therefore a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2) = m(a-b)(b-c)(c-a)$$
Please guide further on how to find this coefficient.

$$a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)$$
$$=ab^2-ac^2+bc^2-a^2b+a^2c-cb^2$$
Now the method to factor cyclic expressions is to arrange the expression with the highest powers of the first variable, i.e:
We take powers of $a$ $$a^2c-a^2b+ab^2-ac^2+bc^2-cb^2$$
$$=a^2(c-b)-a(c^2-b^2)+bc(c-b)$$
$$=(c-b)(a^2-ac-ab+bc)$$
Now we look for powers of $b$,: 
$$=(c-b)(bc-ab-ac+a^2)$$
$$=(c-b)(b(c-a)-a(c-a))$$
$$=(c-b)(c-a)(b-a)$$
$$=(a-b)(b-c)(c-a)$$
Thus, $m$ is $1$.
