$\begin{cases}y(x+y)^2=2\\ 8y(x^3-y^3)=13 \end{cases}$ $ x>y, y>0\Rightarrow x>0$ Let $x=y+h, h>0$ Substituting $x$ by $y+h$ in the fraction obtained by dividing the second equation by the first one leads to: $\Longrightarrow\frac{8h(3y^2+3hy+h^2)}{(2y+h)^2}=\frac{13}{2}$ $\Longrightarrow 16h(3y^2+3hy+h^2)=13(4y^2+h^2+4hy)$ $\Longrightarrow 4y^2+4hy-3h^2=0\Longrightarrow y=\frac{-4h\pm\sqrt{16h^2+48h^2}}{8}$ $\Longrightarrow y= \frac{1}{2}h, y=-\frac{3}{2}h$ (the latter has to be discarded as it produces negative x,y) $\Longrightarrow x=y+h \Longrightarrow x= h+\frac{1}{2}h=\frac{3}{2}h$ $\Longrightarrow \frac{x}{y}=\frac{\frac{3}{2}h}{\frac{1}{2}h}=3$ $\Longrightarrow x=3y$ Substituting $x=3y$ in the first equation results in $16y^3=2\Longrightarrow y=\frac{1}{2}$ In conclusion, $(x,y)=(\frac{3}{2}, \frac{1}{2})$ (sole solution). @ below: désolé pour vous avoir "snipé"

Blastoor wrote: Find all ordered pairs $(x,y)$ of real numbers that satisfy the following system of equations: $$\begin{cases} y(x+y)^2=2\\ 8y(x^3-y^3) = 13. \end{cases}$$ First equation implies $y>0$ and second implies then $x>y>0$ Let $u=x-y>0$ and $v=2y>0$ First equation is $v(u+v)^2=4$ and so $u^2=-2uv+\frac 4v-v^2$ So $u^3=-2u^2v+(\frac 4v-v^2)u$ $=-2v(-2uv+\frac 4v-v^2)+(\frac 4v-v^2)u$ $=u(3v^2+\frac 4v)+2v^3-8$ Second equation is $v(3v^2u+6vu^2+4u^3)=13$ and plugging there the previously got expressions for $u^2$ and $u^3$, we get $u(3v^2+\frac{16}v)+2v^3-8=\frac{13}v$ and so $u=\frac{13+8v-2v^4}{3v^3+16}$ Plugging this in first equation $v(u+v)^2=4$, we get $v^9+12v^6+26v^5+192v^3+624v^2+169v-1024=0$ Not very difficult to see that $v=1$ is a "trivial" root and, factoring $v-1$, we get $(v-1)(v^8+v^7+v^6+13v^5+39v^4+39v^3+231v^2+855v+1024)=0$ The second factor is alwayts $>0$ since $v>0$ and so the only solution is $v=1$ and so $u=\frac{13+8v-2v^4}{3v^3+16}=1$ Hence the unique real solution $\boxed{(x,y)=\left(\frac 32,\frac 12\right)}$

Clearly $y>0$ and $x>y$. Now, setting $t:=x/y$, we get \[ y^3(t+1)^2 = 2\quad\text{and}\quad y^4(t^3-1) = \frac{13}{8}. \]Raising the first equation into fourth and the second into third power, we get \[ 13^3 \cdot (t+1)^8 = 16 \cdot 8^3 \cdot (t^3-1)^3. \]This annoying expression factors as \[ (t - 3) (8192 t^8 + 22379 t^7 + 49561 t^6 + 62591 t^5 + 64741 t^4 + 40433 t^3 + 22843 t^2 + 7013 t + 3463) = 0, \]and as $t>1$, we get $t=3$. With this, we get $y^4 = 1/16$. As $y>0$, we have $y=1/2$ which, in turn, yields $x=3/2$.