Hello, you can help me please ?

Prove that $\displaystyle (x,y) = (2,1)$ is the only solution of the system : $\displaystyle \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.$

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- Apr 1st 2009, 03:18 PMlinda2005System
Hello, you can help me please ?

Prove that $\displaystyle (x,y) = (2,1)$ is the only solution of the system : $\displaystyle \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.$ - Apr 2nd 2009, 06:40 AMNonCommAlg
clearly from the equations we have $\displaystyle x > y> 0.$ let $\displaystyle y=t^2, \ t > 0.$ so from the second equation we get: $\displaystyle x=\frac{3}{t} - t^2$ which with the first equation gives us: $\displaystyle 2t^9 - 9t^6 + 27t^3 + 7t - 27=0. \ \ \ (*)$

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

$\displaystyle f'(z)=18z^8 - 54z^5 + 81z^2 + 7=9z^2((z^3 - 3)^2 + z^6) + 7 \geq 7.$ thus $\displaystyle f'(z) \neq 0, \ \forall z \in \mathbb{R},$ and thereofore it is not possible for $\displaystyle f$ to have more than one real zero by Rolle's theorem. Q.E.D. - Apr 2nd 2009, 08:21 AMDeMath
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.

http://s47.radikal.ru/i115/0904/99/559e3cce7fbat.jpg - Apr 2nd 2009, 04:46 PMNonCommAlg
- Apr 2nd 2009, 06:04 PMDeMath
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 (Happy)(Doh) on Department of Mechanics and Mathematics. As far as I know this problem to fully solve only two entrants.