Solve the following system of equations:

$$
x^{3}+x^{3} y^{3}+y^{3}=12, \quad x+x y+y=0
$$
Transform the left side of the first equation:

$$
\left(x^{3}+y^{3}\right)+x^{3} y^{3}=(x+y)^{3}-3 x y(x+y)+x^{3} y^{3}=12
$$

and substitute the expression for \(x+y\) obtained from the second equation:

$$
3 x^{2} y^{2}=12, \quad x y= \pm 2, \quad \text { and thus } \quad x+y=\mp 2 .
$$

Now we have the system of equations

$$
x y=2, \quad x+y=-2
$$

whose solutions are the roots \(u_{1}, u_{2}\) of the equation

$$
u^{2}+2 u-2=0
$$

taken in any order:

$$
x_{1}=u_{1}=-1+\sqrt{3}, \quad y_{1}=u_{2}=-1-\sqrt{3} ; \quad x_{2}=y_{1}, \quad y_{2}=x_{1}
$$

and from the system of equations \(x y=-2, x+y=2\) similarly

$$
x_{3}=1+\sqrt{3}, \quad y_{3}=1-\sqrt{3} ; \quad x_{4}=y_{3}, \quad y_{4}=x_{3}
$$

Ferenc Czina (Makó, József A. g. II. o. t.)

See also the 1964/2 79. page for corrections. [The editor.]