<s>[INST] Factorise $Σa^{2}(b^{4}-c^{4})$ 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).
 [/INST] 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 $a=b=c$.
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