Question

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

Answer 1

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