已知x=2a-b-c,y=2b-a-c,z=2c-a-b化简(b-c)x+(c-a)y+(a-b)z

x+y+z=2a-b-c+2b-a-c+2c-a-b=0
(b-c)x+(c-a)y+(a-b)z①
=a(z-y)+b(x-z)+c(y-z)
=a(x+2z)+b(y+2x)+c(z+2y)
=(a+2b)x+(b+2c)y+(c+2a)②
∴对比①②可得a+b+c=0
∴(b-c)x+(c-a)y+(a-b)z
=(b-c)(2a-b-c)+(c-a)(2b-a-c)+(a-b)(2c-a-b)
=(b-c)3a+(c-a)3b+(a-b)3c
=0