Ta có:

ab(a – b) + bc(b – c) + ca(c – a)

= ab(a – b) + b²c – bc² + c²a – a²c

= ab(a – b) + (b²c – ca²) + (c²a – c²b)

= ab(a – b) + c(b² – a²) + c²(a – b)

= ab(a – b) – c(a – b)(a + b) + c²(a – b)

= (a – b)(ab – ca – cb + c²)

= (a – b)[(ab – cb) + (–ca + c²)]

= (a – b)[b(a – c) + c(c – a)]

= (a – b)(a – c)(b – c).