Show that for all real numbers $x,y,z$ such that $x + y + z = 0$ and $xy + yz + zx = -3$, the expression $x^3y + y^3z + z^3x$ is a constant.
Problem
Source: CRMO 2012 Region 2 p6
Tags: algebra, identity, algebraic identities
DerJan
30.09.2018 20:16
Note that $$x^3y+y^3z+z^3x=(x^2y+y^2z+z^2x+xyz)(x+y+z)-(xy+yz+zx)^2=-9.$$
HoRI_DA_GRe8
12.08.2021 09:32
DerJan wrote: Note that $$x^3y+y^3z+z^3x=(x^2y+y^2z+z^2x+xyz)(x+y+z)-(xy+yz+zx)^2=-9.$$ I wrote $x=-y-z$ multiplied with $x^2y$ on both sides,then added similar identities and got $x^3y+y^3z+z^3x=-(xy+yz+zx)^2-3(x^2yz+y^2zx+z^2xy)=-(xy+yz+zx)^2-3xyz(x+y+z)=-9-0=-9$ and hence done
OlympusHero
13.08.2021 07:18
DerJan wrote: Note that $$x^3y+y^3z+z^3x=(x^2y+y^2z+z^2x+xyz)(x+y+z)-(xy+yz+zx)^2=-9.$$ How did you find this factorization?
lifeismathematics
29.09.2022 20:09
$xy+yz+zx=-3, xy+z(-z)=-3, xy=z^2-3\\
\sum_{cyc} x^3y=\sum_{cyc} x^2(z^2-3) \implies \sum_{cyc} x^2y^2-3(\sum x^2)$
now notice $(\sum xy)^2=\sum x^2y^2-2xyz(x+y+z) \implies \sum x^2y^2=9$
so $\boxed{\sum_{cyc} x^3y=9-18=-9}$
bal09
31.05.2024 10:55
using vieta solutions we can find answer i had made an attachment in that you can see solution
Attachments:
Jishnu4414l
11.09.2024 08:10
Note that $x,y,z$ are roots of the equation in $a$; $a^3-3a-xyz=0$ From the above fact, we have the equation, $x^3-3x-xyz=0\implies yx^3-3xy-xy^2z=0$ Analogously, write the other two equations containing $yx^3, zy^3$ and $xz^3$, and add them. We get, $yx^3+zy^3+xz^3=3(xy+yz+zx)+xyz(x+y+z)=-9$, and we are done!