试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)

有这样的公式：a^3+b^3+c^2-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca) 左边减右边,证明：（x+y-2z)^3+(y+z-2x)^3+(z+x-2y)^3-3（x+y-2z)(y+z-2x)(z+x-2y) =[（x+y-2z)+(y+z-2x)+(z+x-2y)][（x+y-2z)^2+(y+z-2x)^2+(z+x-2y)^2-(x+y-2z)(y+z-2x)- (y+z-2x)(z+x-2y)-(z+x-2y)(x+y-2z)] =0 所以（x+y-2z)^3+(y+z-2x)^3+(z+x-2y)^3=3（x+y-2z)(y+z-2x)(z+x-2y) 回答人：潇湘诗社 ☆國士無雙卍 有疑问欢迎追问,

x-z=a,y-z=b,x-y=c 左边=(X+Y-2Z)^3+ (Y+Z-2X)^3 +（Z+X-2Y)^3 =(a+b)^3-(c+a)^3+(c-b)^3 =a^3+3a^2b+3ab^2+b^3-(a^3+3a^2c+3ac^2+c^3)+c^3-3c^2b+3cb^2-b^3 =3a^2b+3ab^2-3a^2c-3ac^2-3c^2b+3cb^2 =3(a^2b+ab^2-a^2c-ac^2-c^2b+cb^2) =3[a^2(b-c)+a(b+c)(b-c)+bc(b-c)] =3(b-c)(a^2+ab+ac+bc) =3(b-c)(a+c)(a+b) =右边.