I have to solve this system of equations for real $x$ and $y$:
$2x^3+4=x^2(y+3)$
$2y^3+4=y^2(x+3)$
I noticed that they are symetric and, assuming that $x=y$, I got two solutions: $x=y=-1$ or $x=y=2$, which are correct (I checked with the answers), however, I do not know how to show that they are the only solutions. Please, help me.
$(x,y)=(2,2),(-1,-1)$ are the only real solutions.
In the following, suppose that $x\not=y$.
Let $x+y=\alpha,xy=\beta$. Then we have $$\begin{align}&(2x^3+4)+(2y^3+4)=x^2(y+3)+y^2(x+3)\\&\iff 2(x^3+y^3)+8=xy(x+y)+3(x^2+y^2)\\&\iff 2(x+y)((x+y)^2-3xy)+8=xy(x+y)+3((x+y)^2-2xy)\\&\iff 2\alpha(\alpha^2-3\beta)+8=\alpha\beta +3(\alpha^2-2\beta)\tag1\end{align}$$ and $$\begin{align}&(2x^3+4)-(2y^3+4)=x^2(y+3)-y^2(x+3)\\&\iff 2(x^3-y^2)=xy(x-y)+3(x-y)(x+y)\\&\iff 2(x-y)(x^2+xy+y^2)=(x-y)(xy+3(x+y))\\&\iff (x-y)(2\alpha^2-2\beta-\beta-3\alpha)=0\\&\Rightarrow \beta=\frac{2\alpha^2-3\alpha}{3}\tag2\end{align}$$
So, from $(1)(2)$, $$2\alpha(\alpha^2-(2\alpha^2-3\alpha))+8=\alpha\cdot\frac{2\alpha^2-3\alpha}{3} +3\left(\alpha^2-2\cdot\frac{2\alpha^2-3\alpha}{3}\right)$$ $$\iff 4\alpha^3-12\alpha^2+9\alpha -12=0\tag3$$
By the way, since $x,y$ are the roots of $t^2-\alpha t+\beta=0$, if $\alpha^2-4\beta\lt 0$, i.e. if $$\alpha\lt 0\quad\text{or}\quad \alpha\gt 2.4$$ then there is no such $(x,y)\in\mathbb R$.
Now you can easily see that $(3)$ has only one real root $\alpha$ such that $2.5\lt\alpha\lt 3$.
Hence, there are no other real solutions.