System of equations

[ins-]

Solve (in reals) the system:
$x^2y+y^2z+z^2x=23$
$xy^2+yz^2+zx^2=25$
$x^3+y^3+z^3=36$

(Оффтоп)

The following identities might be useful:
$12a^2b^2c^2+(a^3+b^3+c^3)^2+2((a^3+b^3+c^3)abc-(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2))=-2(ab+bc+ca)(a^2b^2+b^2c^2+c^2a^2-a^2bc-ab^2c-abc^2)$
$a^2b+b^2c+c^2a+ab^2+bc^2+ca^2+3abc=(a+b+c)(ab+bc+ca)$
$(a+b+c)^3-6abc=const$

[arqady]

ins- в сообщении #1114719 писал(а):

Solve (in reals) the system:
$x^2y+y^2z+z^2x=23$
$xy^2+yz^2+zx^2=25$
$x^3+y^3+z^3=36$

Пусть $x+y+z=3u$, $xy+xz+yz=3v^2$, где $v^2$ может быть отрицательным, и $xyz=w^3$
Тогда $x^3+y^3+z^3=36$ даёт $9u^3-9uv^2+w^3=12$, а после сложения первых двух уравнений получаем $3uv^2-w^3=16$ откуда
$v^2=\frac{9u^3-28}{6u}$ и $w^3=4.5u^3-30$.
Вычитая из первого уравнения второе, получаем $(x-y)(x-z)(y-z)=-2$, что после возведения в квадрат даёт
$27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=4$ и после подстановки выражений для $v^2$ и $w^3$ получаем $(u-2)(u^2+2u+4)(243u^6-2268u^3+2744)=0$.
$u=2$ даёт тройку $(1,2,3)$ и её циклические перестановки, а вот два других значения $u$ особой радости, похоже, не доставляют.

[scwec]

Вот что дает Maple. Радости точно нет.

{x = 3, y = 1, z = 2}, {x = 2, y = 3, z = 1}, {x = 1, y = 2, z = 3},
{x = RootOf(_Z+1+_Z^2), y = 2*RootOf(_Z+1+_Z^2), z = 3*RootOf(_Z+1+_Z^2)},
{x = 2*RootOf(_Z+1+_Z^2), y = 3*RootOf(_Z+1+_Z^2), z = RootOf(_Z+1+_Z^2)},
{x = 3*RootOf(_Z+1+_Z^2), y = RootOf(_Z+1+_Z^2), z = 2*RootOf(_Z+1+_Z^2)},

{x = RootOf(_Z^3-RootOf(27*_Z^6-1944*_Z^5+36018*_Z^4+312012*_Z^3-14409873*_Z^2+100605348*_Z-61162984)),
y = -(1/2502085875016*(629280438608-11027163380386*RootOf(27*_Z^6-1944*_Z^5+36018*_Z^4+312012*_Z^3-14409873*_Z^2+100605348*_Z-61162984)[phone]*RootOf(27*_Z^6-1944*_Z^5+36018*_Z^4+312012*_Z^3-14409873*_Z^2+100605348*_Z-61162984)^2+9295210653*RootOf(27*_Z^6-1944*_Z^5+36018*_Z^4+312012*_Z^3-14409873*_Z^2+100605348*_Z-61162984)^3-3713798439*RootOf(27*_Z^6-1944*_Z^5+36018*_Z^4+312012*_Z^3-14409873*_Z^2+100605348*_Z-61162984)^4+82283895*RootOf(27*_Z^6-1944*_Z^5+36018*_Z^4+312012*_Z^3-14409873*_Z^2+100605348*_Z-61162984)^5))*RootOf(_Z^3-RootOf(27*_Z^6-1944*_Z^5+36018*_Z^4+312012*_Z^3-14409873*_Z^2+100605348*_Z-61162984)),
z = (1/2502085875016*(20272377039760-19060976927914*RootOf(27*_Z^6-1944*_Z^5+36018*_Z^4+312012*_Z^3-14409873*_Z^2+100605348*_Z-61162984)[phone]*RootOf(27*_Z^6-1944*_Z^5+36018*_Z^4+312012*_Z^3-14409873*_Z^2+100605348*_Z-61162984)^2+20658047055*RootOf(27*_Z^6-1944*_Z^5+36018*_Z^4+312012*_Z^3-14409873*_Z^2+100605348*_Z-61162984)^3-6043870989*RootOf(27*_Z^6-1944*_Z^5+36018*_Z^4+312012*_Z^3-14409873*_Z^2+100605348*_Z-61162984)^4+130444281*RootOf(27*_Z^6-1944*_Z^5+36018*_Z^4+312012*_Z^3-14409873*_Z^2+100605348*_Z-61162984)^5))*RootOf(_Z^3-RootOf(27*_Z^6-1944*_Z^5+36018*_Z^4+312012*_Z^3-14409873*_Z^2+100605348*_Z-61162984))}

[ins-]

arqady
Thank you for the solution! Very impressive!
I don't know what is the level of difficulty of this problem or if it is beautiful. I created it today. Problem 10.2 from here: http://www.botik.ru/~duzhin/olymp/go00sol.ru.html inspired me. It seems it is not suitable for a competition and it is hard to be solved without a computer.
"$v^2$ может быть отрицательным" - I suppose it means that $v$ is a complex variable. Am I correct?

[arqady]

ins- в сообщении #1114774 писал(а):

I suppose it means that $v^2$ is a complex variable. Am I correct?

No, you are wrong. I said only that $v^2$ can be a negative real number.

[ins-]

I was surprised, because I know that square of something is always a positive number.

[ins-]

In a little modified way I calculated that the only possible values for $xyz$ are $6$, $-9-7 \sqrt {\frac{13}{3}}$, $-9+7 \sqrt {\frac{13}{3}}$. Can, using this, the exact values for $x$, $y$, $z$ be found? At least for xyz=6 they are nice.