
By §~8, any number has three cube roots, two of which are the products
of the remaining one by the imaginary cube roots of unity:
\[ % Lining up parallel structures guards against typos.
\tag{8}
\omega   = -\tfrac{1}{2} + \tfrac{1}{2} \sqrt{3}i,\qquad
\omega^2 = -\tfrac{1}{2} - \tfrac{1}{2} \sqrt{3}i.
\]
We can choose particular cube roots
\[
\tag{9}
A = \sqrt[3]{-\frac{q}{2} + \sqrt{R}},\qquad
B = \sqrt[3]{-\frac{q}{2} - \sqrt{R}},
\]
such that $AB = -p/3$, since the product of the numbers under the cube
root radicals is equal to $(-p/3)^3$. Hence the six values of~$z$ are
\[
A,\quad \omega A,\quad \omega^2 A,\quad
B,\quad \omega B,\quad \omega^2 B.
\]
These can be paired so that the product of the two in each pair is $-p/3$.
\[
AB = -\frac{p}{3}, \qquad
\omega   A \cdot \omega^2 B = -\frac{p}{3}, \qquad
\omega^2 A \cdot \omega   B = -\frac{p}{3}.
\]
Hence with any root~$z$ is paired a root equal to $-p/(3z)$. By~$(5)$, the sum
of the two is a value of~$y$. Hence the three values of~$y$ are
\[
\tag{10}
y_1 = A + B,\qquad
y_2 = \omega   A + \omega^2 B,\qquad
y_3 = \omega^2 A + \omega   B.
\]

It is easy to verify that these numbers are actually roots of~$(2)$. For
example, since $\omega^3 = 1$, the cube of $y_2$ is
\[
A^3 + B^3 + 3\omega A^2 B + 3\omega^2 AB^2
  = -q - p(\omega A + \omega^2 B)
  = -q - py_2,
\]
by~$(9)$ and $AB = -p/3$.

The numbers~$(10)$ are known as \emph{Cardan's formulas} for the roots of a
reduced cubic equation~$(2)$. The expression $A + B$ for a root was first
published by Cardan in his \emph{Ars Magna} of~1545, although he had obtained
it from Tartaglia under promise of secrecy.

% [** F1: \small?]
\textsc{Example}.\quad Solve $y^3 - 15y - 126 = 0$.

\textit{Solution}.\quad The substitution~$(5)$ is here $y = z + 5/z$.  We get
\[
z^6 - 126z^3 + 125 = 0,\qquad  z^3 = 1 \text{ or } 125.
\]
The pairs of values of~$z$ whose product is $5$ are $1$ and~$5$, $\omega$ and $5\omega^2$, $\omega^2$ and $5\omega$. Their
sums $6$, $\omega + 5\omega^2$, and $\omega^2 + 5\omega$ give the three roots.
% [** F1: End of \small]

\begin{center}
\textbf{EXERCISES}
\end{center}

% [** F1: \small?]
Solve the equations:
\begin{itemize}

\item[1.] $y^3 - 18y + 35 = 0$.

\item[2.] $x^3 + 6x^2 + 3x + 13 %[**unclear][** F1: Looks like 18...?]
 = 0$.

\item[3.] $y^3 - 2y + 4 = 0$.

\item[4.] $28x^3 + 9x^2 - 1 = 0$.
\end{itemize}
% [** F1: End of \small]
