1. ## System

Hello, you can help me please ?

Prove that $(x,y) = (2,1)$ is the only solution of the system : $\quad \large (x,y) \in {\mathbb R}^2 ,\quad \left\{\begin{array}{l} y(x^3 - y^3) = 7 \\
y(x + y)^2 = 9 \end{array} \right.$

2. Originally Posted by linda2005

Hello, you can help me please ?

Prove that $(x,y) = (2,1)$ is the only solution of the system : $\quad \large (x,y) \in {\mathbb R}^2 ,\quad \left\{\begin{array}{l} y(x^3 - y^3) = 7 \\
y(x + y)^2 = 9 \end{array} \right.$
clearly from the equations we have $x > y> 0.$ let $y=t^2, \ t > 0.$ so from the second equation we get: $x=\frac{3}{t} - t^2$ which with the first equation gives us: $2t^9 - 9t^6 + 27t^3 + 7t - 27=0. \ \ \ (*)$

obviously $t=1$ satisfies $(*).$ we'll show that that's the only real solution of $(*).$ to see this, let $f(z)=2z^9 - 9z^6 + 27z^3 + 7z - 27, \ \ z \in \mathbb{R}.$ then:

$f'(z)=18z^8 - 54z^5 + 81z^2 + 7=9z^2((z^3 - 3)^2 + z^6) + 7 \geq 7.$ thus $f'(z) \neq 0, \ \forall z \in \mathbb{R},$ and thereofore it is not possible for $f$ to have more than one real zero by Rolle's theorem. Q.E.D.

3. Originally Posted by linda2005
Hello, you can help me please ?

Prove that $(x,y) = (2,1)$ is the only solution of the system : $\quad \large (x,y) \in {\mathbb R}^2 ,\quad \left\{\begin{array}{l} y(x^3 - y^3) = 7 \\
y(x + y)^2 = 9 \end{array} \right.$
View the solution of your problems, for example, here (Problem 12, pages 29-30). You need a special program to view this file.

http://www.lix.polytechnique.fr/Labo...ardi/M00-06.ps

These two pages of evidence from the file.

4. Originally Posted by DeMath
View the solution of your problems, for example, here (Problem 12, pages 29-30). You need a special program to view this file.

http://www.lix.polytechnique.fr/Labo...ardi/M00-06.ps

These two pages of evidence from the file.

i'm just wondering if there's anybody who thinks this solution is better than mine!

5. Originally Posted by NonCommAlg
i'm just wondering if there's anybody who thinks this solution is better than mine!
Of course your solution is better, elegant. Just by chance I found it on the Internet and suggested as a possible solution.

Incidentally, this problem was proposed by professor of my university for entrance examinations on Department of Mechanics and Mathematics. As far as I know this problem to fully solve only two entrants.