>>467
(a-c)^3+(b-c)^3+(-a-b+2c)^3
=(a+b-2c)((a-c)^2-(a-c)(b-c)+(b-c)^2)+(-a-b+2c)^3
=(a+b-2c)((a-c)^2-(a-c)(b-c)+(b-c)^2-(a+b-2c)^2)
=(a+b-2c)((a-c)^2-(a-c)(b-c)+(b-c)^2+(a+b-2c)^2)
=(a+b-2c)(-3ab+3ac+3bc-3c^2)
=3(a+b-2c)(a-c)(c-b)