丁寧に書いてみます(=^・^=)

[解]

\(a^{4}\)+\(b^{4}\)+\(c^{4}\)-2\(b^{2}\)\(c^{2}\)-2\(c^{2}\)\(a^{2}\)-2\(a^{2}\)\(b^{2}\)

=\(a^{4}\)-2(\(b^{2}\)+\(c^{2}\))\(a^{2}\)+\(b^{4}\)+\(c^{4}\)-2\(b^{2}\)\(c^{2}\)

=\(a^{4}\)-2(\(b^{2}\)+\(c^{2}\))\(a^{2}\)+(\(b^{2}\)-\(c^{2}\)\()^{2}\)

=\(a^{4}\)+2(\(b^{2}\)-\(c^{2}\))\(a^{2}\)+(\(b^{2}\)-\(c^{2}\)\()^{2}\)-4\(a^{2}\)\(b^{2}\)

=(\(a^{2}\)+\(b^{2}\)-\(c^{2}\)\()^{2}\)-4\(a^{2}\)\(b^{2}\)

=(\(a^{2}\)+2ab+\(b^{2}\)-\(c^{2}\))(\(a^{2}\)-2ab+\(b^{2}\)-\(c^{2}\))

={(a+b\()^{2}\)-\(c^{2}\)}{(a-b\()^{2}\)-\(c^{2}\)}

=(a+b+c)(a+b-c)(a-b+c)(a-b-c)

ここでやめて十分ですが、対称性を重んずるなら以下

=-(a+b+c)(a+b-c)(a-b+c)(-a+b+c)
または
=(a+b+c)(-a-b+c)(-a+b-c)(a-b-c)

与式＝\(a^{4}\)-2(\(b^{2}\)+\(c^{2}\))\(a^{2}\)+\(b^{4}\)+\(c^{4}\)-2\(b^{2}\)\(c^{2}\)
＝\(a^{4}\)-2(\(b^{2}\)+\(c^{2}\))+(\(b^{2}\)-\(c^{2}\)\()^{2}\)
＝{\(a^{2}\)-(\(b^{2}\)+\(c^{2}\))}^2-(\(b^{2}\)+\(c^{2}\)\()^{2}\)+(\(b^{2}\)-\(c^{2}\)\()^{2}\)
＝(\(a^{2}\)-\(b^{2}\)-\(c^{2}\)\()^{2}\)-4\(b^{2}\)\(c^{2}\)
＝{(\(a^{2}\)-\(b^{2}\)-\(c^{2}\))+2bc}{(\(a^{2}\)-\(b^{2}\)-\(c^{2}\))-2bc}
＝{\(a^{2}\)-(b-c\()^{2}\)}{\(a^{2}\)-(b+c\()^{2}\)}
＝{a-(b-c)}{a+(b-c)}{a-(b+c)}{a+(b+c)}
＝(a-b+c)(a+b-c)(a-b-c)(a+b+c)・・・（答）