24.2. Find the roots of the equation $x^{3}+p x+q=0$, representing them in the form $x=\sqrt[3]{\alpha}+\sqrt[3]{\beta}$ and finding expressions for $\alpha$ and $\beta$.
24.2. The equality must hold

$$
x^{3}=\alpha+\beta+3 \sqrt[3]{\alpha \beta}(\sqrt[3]{\alpha}+\sqrt[3]{\beta})=\alpha+\beta+3 \sqrt[3]{\alpha \beta} x
$$

Therefore, the numbers $\alpha$ and $\beta$ must be chosen so that the equalities $3 \sqrt[3]{\alpha \beta}=-p$ and $\alpha+\beta=-q$ are satisfied. From these equalities, it follows that $\alpha \beta=-\frac{p^{3}}{27}$ and $\alpha+\beta=-q$. Thus, for the numbers $\alpha$ and $\beta$,

Chapter 24. Equations Solvable in Radicals 290

a quadratic equation is obtained. Its roots have the form

$$
\alpha, \beta=-\frac{q}{2} \pm \sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}
$$

The formula $x=\sqrt[3]{\alpha}+\sqrt[3]{\beta}$ gives 9 different values.

To obtain a formula that gives 3 values, one can use the relation $\sqrt[3]{\alpha} \sqrt[3]{\beta}=-p / 3$. This formula has the form

$$
x=\sqrt[3]{\alpha}-\frac{p}{\sqrt[3]{\alpha}}
$$

where $\alpha$ is found using formula (1). In this case, $x$ does not depend on the choice of the sign in front of the radical in formula (1). It is not difficult to verify that the obtained formula gives values of $x$ that are roots of the equation $x^{3}+p x+q=0$.