1. Find all real numbers $x$ and $y$ that satisfy the equations

$$
\begin{aligned}
x^{3}+8 y^{3} & =x+2 y \\
2 x^{2} y+4 x y^{2} & =x+2 y
\end{aligned}
$$
II/1. The equations can be rewritten in the form

$$
\begin{aligned}
(x+2 y)\left(x^{2}-2 x y+4 y^{2}\right) & =x+2 y \\
2 x y(x+2 y) & =x+2 y
\end{aligned}
$$

Obviously, any pair of numbers $x$ and $y$ that satisfy the relation $x+2 y=0$ solves the equations. Let now $x+2 y \neq 0$. Then we can divide by $x+2 y$ and get $x^{2}-2 x y+4 y^{2}=1$ and $2 x y=1$, so

$$
(x+2 y)^{2}=x^{2}+4 x y+4 y^{2}=\left(x^{2}-2 x y+4 y^{2}\right)+6 x y=4
$$

We distinguish two cases, either $x+2 y=2$ or $x+2 y=-2$. In the first case, we get $1=2 x y=2(2-2 y) y=4 y-4 y^{2}$, so $0=4 y^{2}-4 y+1=(2 y-1)^{2}$. It follows that $y=\frac{1}{2}$ and $x=2-2 y=1$. In the second case, $x=-2-2 y$ and thus $1=2 x y=2(-2-2 y) y=-4 y-4 y^{2}$, so $0=(2 y+1)^{2}$. It follows that $y=-\frac{1}{2}$ and $x=-1$.

The equations are therefore satisfied by all real numbers $x$ and $y$ for which $x+2 y=0$, and in addition by $x=-1, y=-\frac{1}{2}$ and $x=1, y=\frac{1}{2}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_16ef8862fb9f11daffb9g-08.jpg?height=74&width=1650&top_left_y=1459&top_left_x=206)

![](https://cdn.mathpix.com/cropped/2024_06_07_16ef8862fb9f11daffb9g-08.jpg?height=58&width=1633&top_left_y=1513&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_16ef8862fb9f11daffb9g-08.jpg?height=54&width=1633&top_left_y=1566&top_left_x=220)

If $x \neq-2 y$, the equations can be divided by ...............................................................

Finding $x+2 y= \pm 2$ (or $x-2 y=0$ or $x^{4}-2 x^{2}+1=0$ ) ................ 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_16ef8862fb9f11daffb9g-08.jpg?height=65&width=1633&top_left_y=1715&top_left_x=220)