Question

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

Answer 1

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

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