1. Find all real numbers $x$ and $y$ for which

$$
x^{3}-y^{3}=7(x-y) \quad \text { and } \quad x^{3}+y^{3}=5(x+y)
$$
II/1. We rewrite the equations as $(x-y)\left(x^{2}+x y+y^{2}\right)=7(x-y)$ and $(x+y)\left(x^{2}-x y+y^{2}\right)=5(x+y)$. If $x=y$, from the second equation we get $2 x^{3}=10 x$, hence $x=0$ or $x= \pm \sqrt{5}$. In the opposite case, we can divide the first equation by $x-y$ and obtain

$$
x^{2}+x y+y^{2}=7
$$

If $x=-y$, from the first equation we get $2 x^{3}=14 x$, hence $x=0$ or $x= \pm \sqrt{7}$. In the opposite case, we can divide the second equation by $x+y$ and obtain

$$
x^{2}-x y+y^{2}=5
$$

The sum of the thus obtained equations gives $x^{2}+y^{2}=6$, and the difference gives $x y=1$, so $(x+y)^{2}=$ $x^{2}+2 x y+y^{2}=8$ and thus $x+y= \pm \sqrt{8}$. Considering that $x=\frac{1}{y}$, we get a quadratic equation for $y$, namely $y^{2} \mp \sqrt{8} y+1=0$, from which it follows that $y= \pm(\sqrt{2}+1)$ or $y= \pm(\sqrt{2}-1)$. In the first case, $x= \pm(\sqrt{2}-1)$, and in the second case, $x= \pm(\sqrt{2}+1)$. All solutions $(x, y)$ are thus the pairs $(0,0),(\sqrt{5}, \sqrt{5}),(-\sqrt{5},-\sqrt{5}),(\sqrt{7},-\sqrt{7}),(-\sqrt{7}, \sqrt{7}),(\sqrt{2}+1, \sqrt{2}-1),(-\sqrt{2}-1,-\sqrt{2}+1)$, $(\sqrt{2}-1, \sqrt{2}+1)$ and $(-\sqrt{2}+1,-\sqrt{2}-1)$.