由于（y-z)²+(x-y)²+(z-x)²=(y+z-2x)²+(x+z-2y)²+(x+y-2z)²,则(y+z-2x)²+(x+z-2y)²+(x+y-2z)²-（y-z)²-(x-y)²-(z-x)²=0=》(y+z-2x+y-z)(y+z-2x-y+z)+...

　　(y+z-2x)²+(x+z-2y)²+(x+y-2z)²-（y-z)²-(x-y)²-(z-x)²=0=》(y+z-2x+y-z)(y+z-2x-y+z)+(x+z-2y+x-z)(x+z-2y-x+z)+(x+y-2z+y-x)(x+y-2z-y+x)=0这是怎么得的？

　　用平方差公式。(y+z-2x)²+(x+z-2y)²+(x+y-2z)²-（y-z)²-(x-y)²-(z-x)²=[(y+z-2x)²-（y-z)²]+[(x+z-2y)²-(x-y)²]+[(x+y-2z)²-(z-x)²]=[(y+z-2x+y-z)+(y-z)][(y+z-2x)-(y-z)]+[(x+z-2y)+(x-z)][(x+z-2y)-(x-z)]+[(x+y-2z)+(y-x)][(x+y-2z)-(y-x)]=(y+z-2x+y-z)(y+z-2x-y+z)+(x+z-2y+x-z)(x+z-2y-x+z)+(x+y-2z+y-x)(x+y-2z-y+x)

　　额！！@……麻烦你能不能把全部完整过程都写下来？？？

　　由于（y-z)²+(x-y)²+(z-x)²=(y+z-2x)²+(x+z-2y)²+(x+y-2z)²，则(y+z-2x)²+(x+z-2y)²+(x+y-2z)²-（y-z)²-(x-y)²-(z-x)²=0=》(y+z-2x)²+(x+z-2y)²+(x+y-2z)²-（y-z)²-(x-y)²-(z-x)²=0=》[(y+z-2x)²-（y-z)²]+[(x+z-2y)²-(x-y)²]+[(x+y-2z)²-(z-x)²]=0=》[(y+z-2x+y-z)+(y-z)][(y+z-2x)-(y-z)]+[(x+z-2y)+(x-z)][(x+z-2y)-(x-z)]+[(x+y-2z)+(y-x)][(x+y-2z)-(y-x)]=0=》（2y-2x)(2z-2x)+(2x-2y)(2z-2y)+(2y-2z)(2x-2z)=0=》4(y-x)(z-x)+4(x-y)(z-y)+4(y-z)(x-z)=0=》2[(y-x)(z-x)+(x-y)(z-y)]+2[(y-x)(z-x)+(y-z)(x-z)]+2[(x-y)(z-y)+(y-z)(x-z)]=0=》2(y-x)(z-x-z+y)+2(z-x)(y-x-y+z)+2(z-y)(x-y-x+z)=0=》2(y-x)²+2(z-x)²+2(z-y)²=0=》所以y-x=0,z-x=0,z-y=0则x=y=z所以原式=(x²+1)³/(x²+1)³=1