Exercise 1.1.4

Proof. The solutions of the equation $y^3-3y=0$ are $0,\pm \sqrt{3}$.

Here $p∕3=-1$ and $q∕2=0$, so the Cardan’s formulas give, with the choice $\sqrt{-1}=i$,

As ${(-i)}^3=i$, we can choose $z_1=\sqrt[3]{i}=-i$, which implies the value $z_2=\sqrt[3]{-i}=i$, since $z_1{z}_2=-p∕3=1$. The Cardan’s formulas give

So $y_2=-\mathit{i\omega }+i{\omega }^2=-\mathit{i\omega }+i(-1-\omega )=-i(2\omega +1)$.

As $\omega =-\frac{1}{2}+i\frac{\sqrt{3}}{2}$, so $2\omega +1=i\sqrt{3}$, $y_2=\sqrt{3}$, and $y_3=-y_2=-\sqrt{3}$.

The Cardan’s formulas give the three real roots of $y^3-3y=0$, with intermediate steps in the complex field. □