x+y+z=3 x^2+y^2+z^2=23 x^3+y^3+z^3=45 求xyz=?

来源：百度知道 编辑：UC知道 时间：2024/05/14 23:32:15

(x+y+z)^2-(x^2+y^2+z^2)
=2(xy+yz+zx)=9-23=-14
xy+yz+zx=-7

x+y=3-z
y+z=3-x
z+x=3-y

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

3*23=45+3*(-7)-3xyz

xyz=-15

(x+y+z)^2-(x^2+y^2+z^2)=2(xy+yz+zx),
xy+yz+zx=[(x+y+z)^2-(x^2+y^2+z^2)]/2=(9-23)/2=-7

x^3+y^3+z^3-3xyz
=(x+y)^3-3xy(x+y)+z^3-3xyz
=(x+y+z)^3-3xy(x+y)-3(x+y)z(x+y+z)-3xyz
=(x+y+z)^3-3xy(x+y+z)-3(x+y)z(x+y+z)
=(x+y+z)[(x+y+z)^2-3xy-3(x+y)z]
=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
=3*[23-(xy+yz+zx)]
=3*(23+7)
=90
3xyz=x^3+y^3+z^3-90=45-90=-45
xyz=-15