Determine the coefficients of the following cubic polynomial

$$
f(x) \equiv A x^{3}+B x^{2}+C x+D
$$

if it is known that when $x=1, \quad f(1)=0$, and

$$
\begin{gathered}
x=\frac{2}{3}, \quad f\left(\frac{2}{3}\right)=-4 \quad \text { and finally when } \\
x=\frac{4}{5}, \quad f\left(\frac{4}{5}\right)=-\frac{16}{5}
\end{gathered}
$$
From the conditions, it follows that:

$$
8 A+12 B+18 C+27 D=-108
$$

$$
644 A+80 B+100 C+125 D=-400
$$

From 1):

$$
A=-(B+C+D)
$$

Substituting this into 2) and 3):

$$
4 B+10 C+19 D=-108
$$

$$
16 B+36 C+61 D=-400
$$

Eliminating $B$ from 4) and 5):

$$
4 C+15 D=-32
$$

from which

$$
C=-8-4 D+t
$$

where

$$
4 t=D
$$

and thus

$$
C=-8-15 t
$$

where

$$
4 t=D
$$

Substituting this value into 4), we get:

$$
\begin{gathered}
4 B=-28+74 t \\
B=-7+18 t+k
\end{gathered}
$$

where

$$
2 k=u
$$

and thus finally $A=15-k 15, B=-7+k 37, C=-8-k 30$ and $D=k 8$. The sought function is therefore of the form:

$$
f(x) \equiv\left(15 x^{3}-7 x^{2}-8 x\right)-k\left(15 x^{3}-37 x^{2}+30 x-8\right)
$$

(Aladár Visnya, real gymnasium VIII. class, Pécs.)

The problem was also solved by: Bernát Friedmann, Sátoralja-Ujhely.