题求证:对任意两两不等的三个数a,b,c,(a+b-c)2(a-c)(b-c)+(b+c-a)2(b-a)(c-a)+(c+a-b)2(c-b)(a-b)是常数.这是2012年北京市中学生数学竞赛初二年级试题之一.贵刊2012年8月下P.31给出了一个运算量要求较高的证明,一般的学生不易掌握.本文给出一个一般学生易掌握,且运算量要求不高的换元证法,供学习与欣赏.证明设a=y+z,b=z+x,c=x+y.则原式=4z2(z-x)(z-y)+4y2(y-z)(y-x)2 Question card: for any two or three unequal numbers a, b, c, (a + bc) 2 (ac) (bc) + (b + ca) 2 (ba) (ca) + (c + ab) 2 (cb) (ab) is a constant, which is one of the subjects of the second grade of Beijing’s secondary school mathematics competition in 2012. In August 2012, P.31 gave a proof of high computational requirements, Students are not easy to grasp.This paper gives a change certificate method which is easy to master for general students and does not require a lot of computation for learning and appreciation.We prove that a = y + z, b = z + x, c = x + y. Then the original formula = 4z2 (zx) (zy) + 4y2 (yz) (yx) 2