設(shè)f(x)=x^3-(3yz)x+y^3+z^3
其中y^3+z^3=(y+z)(y^2-yz+z^2),即f(x)=x^3-(3yz)x+(y+z)(y^2-yz+z^2)
嘗試得(x+(y+z))為原式的因式,因?yàn)閒(-(y+z))=-(y+z)^3+(3yz)(y+z)+y^3+z^3=0
于是用大除法計(jì)算(x^3-(3yz)x+y^3+z^3)/(x+(y+z)),得到另一因式為x^2-(y+z)x+y^2+z^2-yz
最后整理得到(x+y+z)(x²+y²+z²-xy-xz-yz)