$$x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(-p)^{2}-2 q=p^{2}-2 q$$

and

$$x_{1}^{3}+x_{2}^{3}=\left(x_{1}+x_{2}\right)^{3}-3 x_{1}^{2} x_{2}-3 x_{1} x_{2}^{2}=\left(x_{1}+x_{2}\right)^{3}-3 x_{1} x_{2}\left(x_{1}+x_{2}\right)=3 p q-p^{3} .$$

If $x_{1}$ and $x_{2}$ are the roots of the equation $a x^{2}+b x+c=0$, then they are also the roots of the equation $x^{2}+\frac{b}{a} x+\frac{c}{a}=0$, thus

$$x_{1}^{2}+x_{2}^{2}=\left(\frac{b}{a}\right)^{2}-2 \frac{c}{a}=\frac{b^{2}-2 a c}{a^{2}}$$

and

$$x_{1}^{3}+x_{2}^{3}=3 \frac{b c}{a^{2}}-\frac{b^{3}}{a^{3}}=\frac{3 a b c-b^{3}}{a^{3}} .$$

The problem was also solved by: Auer Gy., Bauer E., Bendl K., Berger J., Cukor G., Csada I., Dénes M., Ehrenfeld N., Erdős V., Fried E., Füstös P., Kelemen E., Kirchknopf E., Kiss J., Kovács Gy., Lengyel P., Paunz A., Petrik S., Rosenthal M., Sárközy P., Spilzer L., Steiner L., Szőke D., Viola R., Wáhl V., Weisz S.