②ab(a²-b²)+bc(b²-c²)+ca(c²-a²)
=ab(a²-b²)+b³c-bc³+c³a-ca³
=ab(a²-b²)-c(a³-b³)+c³(a-b)
=(a-b)[ab(a+b)-c(a²+ab+b²)+c³]
=(a-b)(a²b+ab²-a²c-abc-b²c+c³)
=(a-b)[a²(b-c)+ab(b-c)-c(b²-c²)]
=(a-b)(b-c)(a²+ab-bc-c²)
=(a-b)(b-c)[a²-c²+b(a-c)]
=(a-b)(a-c)(b-c)(a+b+c)
第一题(a²+b²+c²)和(ab+ac+bc)之间是乘号吧?