已知x+y+z=3,x^2+y^2+z^2=29,x^3+y^3+z^3=45.求xyz及x^4+y^4+z^4

来源：学生作业帮编辑：神马作文网作业帮分类：数学作业时间：2024/11/11 07:11:58

需要用到因式分解x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)
分解过程参见：
http://zhidao.baidu.com/question/46374912.html
(x+y+z)^2-(x^2+y^2+z^2)
=2xy+2xz+2yz
所以xy+xz+yz=[3*3-29]/2=-10
所以
3xyz=x^3+y^3+z^3-(x+y+z)(x^2+y^2+z^2-xy-yz-xz)
=45-3*[29-(-10)]
=-72
xyz=-24

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