已知x³+y³-z³=96,xyz=4,x²+y²+z²-xy+xz+yz=12,x+y-z=?

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已知x³+y³-z³=96,xyz=4,x²+y²+z²-xy+xz+yz=12,x+y-z=?已知x³+y³-z&su

已知x³+y³-z³=96,xyz=4,x²+y²+z²-xy+xz+yz=12,x+y-z=?
已知x³+y³-z³=96,xyz=4,x²+y²+z²-xy+xz+yz=12,x+y-z=?

已知x³+y³-z³=96,xyz=4,x²+y²+z²-xy+xz+yz=12,x+y-z=?
【1】一个因式分解公式:a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca).在上式中,可令:a=x,b=y,c=-z.则:x³+y³-z³+3xyz=(x+y-z)(x²+y²+z²-xy+yz+zx).再把已知条件:x³+y³-z³=96.xyz=4,x²+y²+z²-xy+yz+zx=12代人上式:96+12=12(x+y-z).∴x+y-z=9.

(x²+y²+z²-xy+xz+yz)(x+y-z)=x³+y³-z³+3xyz=108,x²+y²+z²-xy+xz+yz=12所以x+y-z=9。

x³+y³-z³=(x²+y²+z²-xy+xz+yz)*(x+y+z)-3xyz
96=12*(x+y+z)-3*4
x+y+z=9

x3+y3-z3+3xyz,
=[(x+y)3-3x2y-3xy2]-z3+3xyz,
=[(x+y)3-z3]-(3x2y+3xy2-3xyz),
=(x+y-z)[(x+y)2+(x+y)z+z2]-3xy(x+y-z),
=(x+y-z)(x2+2xy+y2+xz+yz+z2-3xy),
=(x+y-z)(x2+y2+z2-xy+xz+yz),

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x3+y3-z3+3xyz,
=[(x+y)3-3x2y-3xy2]-z3+3xyz,
=[(x+y)3-z3]-(3x2y+3xy2-3xyz),
=(x+y-z)[(x+y)2+(x+y)z+z2]-3xy(x+y-z),
=(x+y-z)(x2+2xy+y2+xz+yz+z2-3xy),
=(x+y-z)(x2+y2+z2-xy+xz+yz),
∵x3+y3-z3=96,xyz=4,x2+y2+z2-xy+xz+yz=12,
∴96+3×4=12(x+y-z),
解得x+y-z=9.

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