Another system

[ins-]

Solve (in reals) the system:
$x^2y+y^2z+z^2x=23$
$xy^2+yz^2+zx^2=25$
$xyz=6$

[Sergic Primazon]

ins- в сообщении #1114867 писал(а):

Solve (in reals) the system:
$x^2y+y^2z+z^2x=23$
$xy^2+yz^2+zx^2=25$
$xyz=6$

$ p=x+y+z, q= xy+yz+zx, r =xyz$

$6p^3+q^3=2627, pq=66, r=6 $

1. $ x=1, y=2, z=3 $

2. $ x= \frac{2}{\sqrt[3]6}, y=\frac{3}{\sqrt[3]{6}}, z=\frac{6}{\sqrt[3]{6}}$

[ins-]

This one can be solved in many elegant ways. It was the reason to post it.

[ins-]

I suppose symmetric polynomials cannot be used for solving this system:
$x^2y+y^2z-z^2x=19$
$xy^2+yz^2-zx^2=5$
$xyz=6$
, but it can be solved by the ways I meant. Post 8 and 9 are the illustrations: http://artofproblemsolving.com/communit ... 26p6181502