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Find $(x;y)$ sastisfied that: $$\left\{\begin{matrix}x^3+\frac{2\sqrt[3]{2}(x+y+xy+1)}{\sqrt[3]{14-x^2+265x}}=(x+1)^2+y^3\\y^3+\frac{2\sqrt[3]{2}(x+y+xy+1)}{\sqrt[3]{14-y^2+265y}}=(y+1)^2+x^3\end{matrix}\right.$$

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Many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Julian Kuelshammer Nov 15 '12 at 16:17
Also it might help if you said where you encountered this problem. –  Julian Kuelshammer Nov 15 '12 at 16:17
solve for x or for y?? –  ulead86 Nov 15 '12 at 16:17
The question should be retagged, it has nothing to do with linear algebra. –  Nikola Milinković Nov 15 '12 at 16:27
what have you tried? where are you having trouble? –  robjohn Nov 15 '12 at 16:31

1 Answer 1

By the symmetry of your equations, if $(a, b)$ is a solution to the first equation, then $(b, a)$ will be a solution to the second, so we might expect that something of the form $(a, a)$ will be a simultaneous solution to both. Our guess is correct:

Substituting $x=y=a$ into one of the equations gives us $$ a^3+\frac{2\sqrt[3]{2}(a^2+2a+1)}{\sqrt[3]{14+265a-a^2}}=(a+1)^2+a^3 $$ so $$ \frac{2\sqrt[3]{2}(a^2+2a+1)}{\sqrt[3]{14+265a-a^2}}=(a+1)^2 $$ and as long as $a\ne-1$ we can divide both sides by $(a+1)^2$ to obtain $$ \frac{2\sqrt[3]{2}}{\sqrt[3]{14+265a-a^2}}=1 $$ so we have $$ 2\sqrt[3]{2}=\sqrt[3]{14+265a-a^2} $$ and cubing both sides yields $$ 16=14+265a-a^2 $$ I'll let you continue from there.

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