The geometric place of complex numbers $x+i y$ in the complex plane, whose cube is a real number, needs to be constructed. (The construction should be carried out with a compass and straightedge after setting the unit length.)





Translating the text as requested, while preserving the original line breaks and formatting.
$$
\begin{aligned}
& (x+i y)^{3}=x^{3}+i\left(3 x^{2} y-y^{3}\right)-3 x y^{2}= \\
& =x^{3}-3 x y^{2}+y(\sqrt{3} x+y)(\sqrt{3} x-y) i .
\end{aligned}
$$

This expression is real if

$$
y(\sqrt{3} x+y)(\sqrt{3} x-y)=0
$$

from which (since $y=0$ gives real numbers):

$$
\text { 1. } y=-\sqrt{3} x \text { and } 2 . y=\sqrt{3} x \text {. }
$$

The geometric places we are looking for are therefore lines that pass through the origin and which, since $\operatorname{tg} \alpha=\mp \sqrt{3}$, form angles of 120 and 60 degrees, respectively, with the positive direction of the abscissa axis.

With a compass and ruler, these geometric places can be constructed as follows: On the positive abscissa axis, we mark a point that is one unit away from the origin. We draw a circle with twice this distance from the origin, which intersects the perpendicular raised to the abscissa axis at one unit distance in two points. Connecting these points with the origin, we obtain the desired lines.

(Szmodics Hildegárd, Kaposvár.)

The problem was also solved by: Aczél F., Bartók I., Bayer B., Bogdán G., Dessauer A., Enyedi B., Haar A., Kertész F., König D., Lázár L., Mixich P., Pilczer P., Pintér M., Pivnyik I., Póka Gy., Riesz K., Schlesinger A., Schmidl I., Spitzer V., Steiner M., Sümegi Gy., Tóbiás L.