Get the formula for the root of the equation $x^{3}+p x+q=0$ :

$$
x=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}
$$
According to problem $\underline{61258}$, taking $a=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}, b=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}$, we obtain the equation $x^{3}-a^{3}-b^{3}-$

$3 a b x=0$. According to problem $\underline{61260}$, it has three roots: $a+b, \omega a+\omega^{2} b$ and $\omega^{2} a+\omega b$, where $\omega-$ is a cube root of 1.