Упростить (x^3+y^3-z^3-3xyz)/(x^2+y^2+z^2-xy-xz-yz)

Решите задачу:

$\frac{x^3+y^3+z^3-3xyz}{x^2+y^2+z^2-xy-xz-yz} =A\\\\(x^3+y^3)+(z^3-3xyz)=(x+y)(x^2-xy+y^2)+z(z^2-3xy)=\\\\=(x+y)(x^2-xy+y^2)\underline {+z(x^2-xy+y^2)-z(x^2-xy+y^2)}+\\\\+z(z^2-3xy)=(x^2-xy+z^2)((x+y)+z)+\\\\+z(z^2-3xy-(x^2-xy+y^2))=(x^2-xy+z^2)(x+y+z)+\\\\+z(z^2-3xy-x^2+xy-y^2)=(x^2-xy+y^2)(x+y+z)+\\\\+z(z^2-(x^2+2xy+y^2))=(x^2-xy+y^2)(x+y+z)+\\\\+z(z^2-(x+y)^2)=(x^2-xy+z^2)(x+y+z)+\\\\+z(z-x-y)(z+x+y)=\\\\=(x+y+z)(x^2-xy+y^2+z(z-x-y))=\\\\=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$

$A= \frac{(x+y+z)(x^2+y^2+z^2-xy-xz-yz)}{x^2+y^2+z^2-xy-xz-yz} =x+y+z$