I expanded this cyclic expression then found the three factors and assumed third to be constant but after that I got stuck at this question. The options are
(a)$(a-b)^{2}(b-c)^{2}(c-a)^{2}$
(b)$(a-b)(a+b)(b-c)(b+c)(c-a)(c+a)$
(c)$(a+b)^{2}(b+c)^{2}(c+a)^{2}$
(d)None of these
The answer given is option (b).
On
We can consider $\sum a(b^2-c^2)$ and then make the substitutions $a\rightarrow a^2,b\rightarrow b^2, c\rightarrow c^2$ at the end.
$$\sum a(b^2-c^2)$$
$$=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2$$
We can group by the squares:
$$=a^2(c-b)+ b^2(a-c)+c^2(b-a)$$
Now we force a factor of $(b-a)$ from the first two terms:
$$=a^2c-aab+bba-b^2c+c^2(b-a)$$
$$=a^2c-b^2c+ba(b-a)+c^2(b-a)$$
And then a factor of $(a-b)$ drops out of the first two terms:
$$=c(a+b)(a-b)+ba(b-a)+c^2(b-a)$$
$$=(b-a)\left[-c(a+b)+ba+c^2\right]$$
We can easily see $c=b$ is a zero, so:
$$=(b-a)(c-a)(c-b)$$
$$=(a-b)(b-c)(c-a)$$.
Factorizing the difference of two squares comes up a lot here. Let $u=b^4-c^4,\,v=c^4-a^4$ so $a^4-b^4=-u-v$ and the sum is$$\begin{align}a^2u+b^2v-c^2(u+v)&=(a^2-c^2)u+(b^2-c^2)v\\&=(a^2-c^2)(b^2-c^2)(b^2+c^2)+(b^2-c^2)(c^2-a^2)(c^2+a^2)\\&=(b^2-c^2)(c^2-a^2)(-b^2-c^2+c^2+a^2)\\&=(a^2-b^2)(b^2-c^2)(c^2-a^2),\end{align}$$which is (b). Incidentally, (c) is obviously wrong because it doesn't in general vanish if $a=b=c$.