已知x+y+z=7,xy+yz+zx=14,xyz=8,
3个回答
1、x,y,z可以看作方程a^3-7a^2+14a^2-8=0的跟,可求的x,y,z为1,2,4.带入即可.
2、也可以按照下面的思路
x²+y²+z²=（x+y+z）^2-2(xy+yz+zx)=49-28=21.
x³+y³+z³=(x²+y²+z²)(x+y+z)-x(y²+z²)-y(x²+z²)-z(x²+y²)
98=(x+y+z)(xy+yz+zx)=x^2(y+z)+xyz+y^2(x+z)+xyz+z^2(x+y)+xyz=x^2(y+z)+y^2(x+z)+z^2(x+y)+3*8
x^2(y+z)+y^2(x+z)+z^2(x+y)=74
x³+y³+z³=(x²+y²+z²)(x+y+z)-x(y²+z²)-y(x²+z²)-z(x²+y²)=21*7-74=73
x²y²+y²z²+z²x²=(xy+yz+zx)^2-2(x+y+z)xyz=14^2-2*7*8=84
(x+y)(y+z)(z+x)=2xyz+x^2(y+z)+y^2(x+z)+z^2(x+y)=16+74=90
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