24.3. Solve the equation $x^{3}+p x+q=0$, using the identity

$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x+\omega y+\omega^{2} z\right)\left(x+\omega^{2} y+\omega z\right)$, where $\omega^{2}+\omega+1=0$. (Choose $y$ and $z$ such that $-3 y z=p$ and $\left.y^{3}+z^{3}=q.\right)$
24.3. If $-3 y z=p$, then $y^{3} z^{3}=-p^{3} / 27$. Therefore, $y^{3}$ and $z^{3}$ are the roots of the quadratic equation $t^{2}-q t-\frac{p^{3}}{27}=0$. The roots of this equation are $\frac{q}{2} \pm \sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}$. Choose one of the three values of the cube root $\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}$ as $y$, and choose $-\frac{p}{3 y}$ as $z$; alternatively, this can be stated as follows: choose as $z$ one of the values of the cube root $\sqrt[3]{\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}$ for which $-3 y z=p$.

The equation $x^{3}+p x+q=0$ has the following roots: $x_{1}=-(y+z)$, $x_{2}=-\left(\omega y+\omega^{2} z\right)$ and $x_{3}=-\left(\omega^{2} y+\omega z\right)$.