三元三次方程組(高手請進)

let the roots of equation p^3+ap^2+bp+c=0 are x,y and z

by the relationship between roots and coef, we have

x+y+z=-a , xy+yz+zx=b and xyz=-c

then x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx)=a^2-2b

and x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(z+x)

=(x+y+z)^3-3(-a-x)(-a-y)(-a-z)

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

=-a^3+3a^3+3(-a)a^2+3ba-3c

=3ba-3c-a^3

then we have -a=71 , a^2-2b=1779 and 3ba-c-a^3=46871

=>a=71 , b=1631 and c=12121

then p^3-71p^2+1631p-12121=0

=>(p-23)(p-31)(p-17)=0

so x,y,z are 23 , 31 and 17 respectively.