x^3+y^3-z^3=(x+y-z)^3-3x^y+3x^z-3y^2x+3y^2z+3z^2x+3z^2x+6xyz
=(x+y-z)^3-[xy(3x+3y-3z)-xz(3x+3y-3z)-yz(3x+3y-3z)]-3xyz
=(x+y-z)^3-3(xy-xz-yz)(x+y-z)-3xyz
=(x+y-z)^3-[(x+y-z)^2-(x^2+y^2+z^2-xy+xz+yz)](x+y-z)-3xyz
=(x^2+y^2+z^2-xy+xz+yz)(x+y-z)-3xy
由于x^3+y^3-z^3 (x^2+y^2+z^2-xy+xz+yz) 和xyz都是已知的
所以....
晚安
