：x3+y3-z3=96 xyz=4 x2+y2+z2-xy+yz+xz=12 求x+y-z
问题描述：
：x3+y3-z3=96 xyz=4 x2+y2+z2-xy+yz+xz=12 求x+y-z
答
a³+b³+c³-3abc
=[( a+b)³-3a²b-3ab²]+c³-3abc
=[(a+b)³+c³]-(3a²b+3ab²+3abc)
=(a+b+c)[(a+b)²-(a+b)c+c²]-3ab(a+b+c)
=(a+b+c)(a²+b²+2ab-ac-bc+c²)-3ab(a+b+c)
=(a+b+c)(a²+b²+c²-ab-ac-bc)
令a=x,b=y,c=-z
x³+y³-z³+3xyz=(x+y-z)(x²+y²+z²-xy+ac+bc)
96+3*4=(x+y-z)*12
x+y-z=96