Consider the cubic equation \[x^3+ax^2+bx+c=0.\] According to the previous item, the change of variable \(y=x+a/3\), gives us \begin{equation} \label{eq} y^3+py+q=0,\end{equation} where \[p=b+\frac{a^2}{3}, \hspace{2ex} q=\frac{2a^3}{27}-\frac{ab}{3}+c\]

Introduce the new variable \(z\) by \[y=z-\frac{p}{3z}.\] Substituting into (\ref{eq}), we find the quadratic equation in \(z^3\) \[z^6+qz^3-\frac{p^3}{27}=0,\] with the two solutions \[z_1^3 =-\frac{q}{2}+\frac{1}{2}\sqrt{q^2+4p^3/27}, \hspace{3ex} z_2^3 =-\frac{q}{2}-\frac{1}{2}\sqrt{q^2+4p^3/27}.\] Note that \[z_1^3z_2^3=-\frac{p^3}{27},\] and so \[z_2=-\frac{p}{3z_1}.\] Hence \(z_1+z_2=z_1-\frac{p}{3z_1} =y\) is a solution of equation (\ref{eq}).
Introduce the discriminant of (\ref{eq}) as \[\delta = -4p^3-27q^2.\] We will discuss the three cases \(\delta \gt 0\), \(\delta \lt 0\), \(\delta =0\) separately.

If \(\delta \gt 0\), then \(p\) must be negative, \((|q|/2)\sqrt{27/|p|^3}\lt 1\), and \(z_1^3\), \(z_2^3\) are not real. We can write \[z_1^3=-\frac{q}{2}+\frac{i}{2}\sqrt{\frac{4|p|^3}{27}-q^2} \hspace{3ex} z_2^3 =-\frac{q}{2}-\frac{i}{2}\sqrt{\frac{4|p|^3}{27}-q^2}.\] In polar coordinates, \[z_1^3=r e^{i\theta}, \hspace{3ex} z_2^3=r e^{-i\theta},\] where \[r=\left(\frac{|p|}{3}\right)^{3/2}, \hspace{3ex} \cos\theta = -\frac{q}{2}\sqrt{\frac{27}{|p|^3}}, \hspace{3ex} 0\lt \theta \lt \pi. \] Let \(\omega=e^{2\pi i/3}\). Then we obtain the six solutions for \(z\): \[ \begin{array}{ccc} \displaystyle z_{11}=\sqrt{\frac{|p|}{3}} e^{i\theta/3}, & z_{12}=z_{11}\omega, & z_{13}=z_{11}\omega^2\\[2ex] \displaystyle z_{21}=\sqrt{\frac{|p|}{3}} e^{-i\theta/3}, & z_{22}=z_{21}\omega, & z_{23}=z_{21}\omega^2 \end{array}. \] Computing \(z_{ij}-p/(3z_{ij})\), \(i=1,2\), \(j=1,2,3\), we find three distinct real solutions for equation (\ref{eq}): \[y_1=z_{11}+z_{21}=2\sqrt{\frac{|p|}{3}} \cos\left(\frac{\theta}{3}\right), \hspace{3ex} y_2=z_{12}+z_{23}=2\sqrt{\frac{|p|}{3}} \cos\left(\frac{\theta}{3} +\frac{2\pi}{3}\right), \hspace{3ex} y_3=z_{13}+z_{22}=2\sqrt{\frac{|p|}{3}} \cos\left(\frac{\theta}{3}+\frac{4\pi}{3}\right). \] Since \(0\lt \theta/3 \lt \pi/3\), \(\cos(\theta/3)\gt 1/2\), \(\cos(2\pi/3 +\theta/3) \lt -1/2\), and \(|\cos(4\pi/3 +\theta/3)|\lt 1/2\). In particular, \(y_1,y_2,y_3\) are distinct.

To describe the solutions in terms of the coefficients \(p\) and \(q\), we take \(\left(-\frac{q}{2}+\frac{i}{2}\sqrt{-q^2-4p^3/27}\right)^{1/3}\) to be a specific cube root of \(-\frac{q}{2}+\frac{i}{2}\sqrt{-q^2-4p^3/27}\), and \(\left(-\frac{q}{2}-\frac{i}{2}\sqrt{-q^2-4p^3/27}\right)^{1/3}\) its complex conjugate. Then the three real solutions are \begin{eqnarray*} y_1&=&\left(-\frac{q}{2}+\frac{i}{2}\sqrt{-q^2-4p^3/27}\right)^{1/3}+\left(-\frac{q}{2}-\frac{i}{2}\sqrt{-q^2-4p^3/27}\right)^{1/3}\\[1ex] y_2&=&\omega\left(-\frac{q}{2}+\frac{i}{2}\sqrt{-q^2-4p^3/27}\right)^{1/3}+\omega^2 \left(-\frac{q}{2}-\frac{i}{2}\sqrt{-q^2-4p^3/27}\right)^{1/3}\\[1ex] y_3&=&\omega^2\left(-\frac{q}{2}+\frac{i}{2}\sqrt{-q^2-4p^3/27}\right)^{1/3}+\omega\left(-\frac{q}{2}-\frac{i}{2}\sqrt{-q^2-4p^3/27}\right)^{1/3} \end{eqnarray*}

If \(\delta \lt 0\), then \(z_1^3\) and \(z_2^3\) are both real and distinct, and we take \(z_1\) and \(z_2\) to be their real cube roots. Then \(y_1=z_1+z_2\) is the only real solution, and \(y_2=z_1 \omega+z_2\omega^2\), \(y_3=z_1\omega^2 +z_2\omega\) are a pair of complex conjugate solutions.

In terms of the coefficients, \begin{eqnarray*} y_1&=&\left(-\frac{q}{2}+\frac{1}{2}\sqrt{q^2+4p^3/27}\right)^{1/3}+\left(-\frac{q}{2}-\frac{1}{2}\sqrt{q^2+4p^3/27}\right)^{1/3}\\[1ex] y_2&=&\omega\left(-\frac{q}{2}+\frac{1}{2}\sqrt{q^2+4p^3/27}\right)^{1/3}+\omega^2 \left(-\frac{q}{2}-\frac{1}{2}\sqrt{q^2+4p^3/27}\right)^{1/3}\\[1ex] y_3&=&\omega^2\left(-\frac{q}{2}+\frac{1}{2}\sqrt{q^2+4p^3/27}\right)^{1/3}+\omega\left(-\frac{q}{2}-\frac{1}{2}\sqrt{q^2+4p^3/27}\right)^{1/3} \end{eqnarray*}

If \(\delta =0\), then \(z_1^3=z_2^3\), and \(y_1=2z_1\) is a real solution with multiplicity 1, and \(y_2=2z_1(\omega+\omega^2)=-y_1/2\) is a real solution of multiplicity 2. Note that the only reduced cubic with a single root of multiplicity 3 is \(y^3\), in which case \(y_1=y_2=0\).

In terms of the coefficients, \begin{eqnarray*} y_1&=&2\left(-\frac{q}{2}\right)^{1/3}\\[1ex] y_2&=& \left(\frac{q}{2}\right)^{1/3}\\[1ex] \end{eqnarray*}

Concise derivation of Cardano's formula
Cardano's formula can be derived from Euler's identities \[e^{i\theta}=\cos \theta +i\sin \theta, \hspace{3ex} 2 \cos\theta = e^{i\theta}+e^{-i\theta}\] and the trigonometric identity \[\cos(3\theta)=4\cos^3 \theta-3\cos \theta.\] Here is an outline in the case \(\delta \gt 0\), that is equivalent to \((|q|/2)\sqrt{27/|p|^3}\lt 1\): \[y^3=py+q\] \[x=\frac{y}{2}\sqrt{\frac{3}{p}}, \hspace{3ex}Q=\frac{q}{2}\sqrt{\frac{27}{p^3}}\] \[4x^3-3x=Q\] \[x=\cos\left(\frac{\theta}{3}\right)\] \[Q=4\cos^3\left(\frac{\theta}{3}\right)-2\cos\left(\frac{\theta}{3}\right) = \cos \theta\] \[x=\frac{1}{2}\left(e^{i\theta/3}+e^{-i\theta/3}\right)=\frac{1}{2}\left(\cos\theta +i \sin \theta\right)^{1/3} +\frac{1}{2}\left(\cos\theta -i\sin \theta\right)^{1/3}=\frac{1}{2}\left(Q+i\sqrt{1-Q^2}\right)^{1/3} +\frac{1}{2}\left(Q-i\sqrt{1-Q^2}\right)^{1/3}\] \[y=\left(\frac{q}{2}+\frac{i}{2}\sqrt{4p^3-27q^2}\right)^{1/3}+\left(\frac{q}{2}-\frac{i}{2}\sqrt{4p^3-27q^2}\right)^{1/3}.\]

Ferrari's solution of the reduced quartic
Given a reduced quartic of the form \[x^4+2ax^2=bx+c\] complete the square on the left to get \[(x^2+a)^2=bx+c+a^2.\] Now add \(2t(x^2+a) +t^2\) to both sides to complete the square again: \[(x^2+a+t)^2=2tx^2+bx +c+a^2+2ta+t^2,\] and choose \(t\) so that the discriminant of the quadratic on the right side is zero, so that we get \[(x^2+a+t)^2=\left(x+\frac{b}{4t}\right)^2.\] In order to choose \(t\) as above we need to solve the cubic \[8t^3+16at^2+8(a^2+c)t-b^2=0.\]

Descartes' solution of the reduced quartic
Let \(k\) be a positive solution of the equation (a cubic in \(k^2\)): \[k^6+2qk^4+(q^2-4s)k^2-r^2=0.\] Then \[x^4+qx^2+rx+s=\left(x^2+kx+\frac{1}{2}\left(q+k^2-\frac{r}{k}\right)\right) \left(x^2-kx+\frac{1}{2}\left(q+k^2+\frac{r}{k}\right)\right).\]