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