已知x^3+y^3+z^3=96,xyz=4,x^2+y^2+z^2-xy+xz+yz=12,则x+y-z的值是多少?

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

即,
(x+y-z)(x^2+y^2+z^2-xy+xz+yz)=(x^3+y^3-z^3)+3xyz
(x+y-z)*12=96+3*4
(x+y-z)*12=108
x+y-z=9

题目中x^3+y^3+z^3=96符号是否打错了?应该是x^3+y^3-z^3=96
因为根据上面的算法,刚好用上这几个已知条件

供参考