b3c3(b-c)+c3a3(c-a)+a3b3(a-b)因式分解

Question

b3c3(b-c)+c3a3(c-a)+a3b3(a-b) 因式分解

Answer 1

b3c3(b-c)+c3a3(c-a)+a3b3(a-b)
=b3c3(b-c)+c4a3-c3a4+a4b3-a3b4
=b3c3(b-c)+a3(c4-b4)+a4(b3-c3)
=b3c3(b-c)+a3(c-b)(c+b)(c2+b2)+(a4)(b-c)(b2+bc+c2)
=(b-c)[b3c3-a3(c+b)(c2+b2)+(a4)(b2+bc+c2)]
=(b-c)(b3c3-a3c3-a3b3-a3bc2-a3b2c+a4b2+a4bc+a4c2)
=(b-c)[b3(c3-a3)-a3c2(c-a)+a3b2(a-c)+a3bc(a-c)]
=(b-c)(c-a)[b3(c2+ac+a2)-a3c2-a3b2-a3bc]
=(b-c)(c-a)[c2(b3-a3)+abc(b2-a2)+a2b2(b-a)]
=(b-c)(c-a)(b-a)[c2(b2+ab+a2)+abc(b+a)+a2b2]
=(b-c)(c-a)(b-a)[abc(a+b+c)+a2b2+a2c2+b2c2]
=(a-b)(a-c)(b-c)[abc(a+b+c)+a2b2+a2c2+b2c2]