1. Ta có: b−ca+c−ab+a−bc
=bc(b−c)+ac(c−a)+ab(a−b)abc
=−(a−b)(b−c)(c−a)abc
ab−c+bc−a+ca−b
=a(c−a)(a−b)+b(b−c)(a−b)+c(b−c)(c−a)(a−b)(b−c)(c−a)
=−(a3+b3+c3)+(a+b)(b+c)(c+a)−5abc(a−b)(b−c)(c−a)
=−3abc+(−c)(−a)(−b)−5abc(a−b)(b−c)(c−a)
=−9abc(a−b)(b−c)(c−a)
Suy ra: (b−ca+c−ab+a−bc)(ab−c+bc−a+ca−b)=9