Xét tử số:
$x^3+y^3+z^3-3xyz$
$=x^3+3x^2+3xy^2+y^3+z^3-3xyz-3x^2y-3xy^2$
$=(x^3+3x^2y+3xy^2+y^3)+z^3-(3xyz+3x^2y+3xy^2)$
$=(x+y)^3+z^3-3xy(z+x+y)$
$=(x+y+z)[(x-y)^2-z(x+y)+z^2]-3xy(x+y+z)$
$=(x+y+z)(x^2+2xy+y^2-xz-yz+z^2-3xy)$
$=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$
Xét mẫu số:
$(x-y)^2+(y-z)^2+(z-x)^2$
$=x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2$
$=2x^2+2y^2+2z^2-2xy-2xz-2yz$
$=2(x^2+y^2+z^2-xy-yz-xz)$
Vậy: `\frac{x^3+y^3+z^2-3xyz}{(x-y)^2+(y-z)^2+(z-x)^2}=\frac{(x+y+z)(x^2+y^2+z^2-xy-yz-xz)}{2(x^2+y^2+z^2-xy-yz-xz)}=(x+y+z)/2`
