已知x+y+z=3,X的平方+y的平方+z的平方=19,x的立方+y的立方+z的立方=30,求xyz的值
问题描述：
已知x+y+z=3,X的平方+y的平方+z的平方=19,x的立方+y的立方+z的立方=30,求xyz的值
答
x+y+z=3
x^2+y^2+z^2=19
可得 xy+yz+xz=-5
(x^2+y^2+z^2)(x+y+z)=3xyz+x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2)=-15
(x^2+y^2+z^2)(x+y+z)-(x^3+y^3+z^3)=x(y^2+z^2)+y(x^2+z^2)+z(x^2+y^2)=27
3xyz=-42
xyz=-14
答
2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)
=-10
xy+yz+zx=-5
x^3+y^3+z^3-3xyz
=[(x+y)^3-3xy(x+y)]+z^3-3xyz
=[(x+y)^3+z^3]-3xy(x+y+z)
=[(x+y+z)^3-3(x+y)z(x+y+z)]-3xy(x+y+z)
=(x+y+z)^3-3(xy+yz+zx)(x+y+z)
=(x+y+z)[(x+y+z)^2-3(xy+yz+zx)]
=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
=3*(19-(-5))
所以30-3xyz=72
所以xyz=-14