a-b=x,
b-c=y,
a-c=a-b+b-c=x+y,
则(a-b)³+(b-c)³-(a-c)³
=x³+y³-(x+y)³
=x³+y³-(x³+y³+3x²y+3xy²)
=-(3x²y+3xy²)
=-3xy(x+y)
=3(b-a)(b-c)(a-c)
a-b=x,
b-c=y,
a-c=a-b+b-c=x+y,
则(a-b)³+(b-c)³-(a-c)³
=x³+y³-(x+y)³
=x³+y³-(x³+y³+3x²y+3xy²)
=-(3x²y+3xy²)
=-3xy(x+y)
=3(b-a)(b-c)(a-c)