It's hard to imagine how del Ferro must have felt after discovering the solution to his cubic equation - he was arguably the only person on the planet who knew how to solve equations like this. The part of the video were I show the derivation his cubic formula goes pretty fast, so I've included a proper derivation here.
$$x^3 + cx = d$$
let \(x = u+v\)
$$(u+v)^3 + c(u+v) = d \\(u^2 + 2uv + v^2)(u+v) + c(u+v) = d\\(u^3 + 3u^2v + 3uv^2 + v^3) + c(u+v) = d\\u^3 + v^3 + 3uv(u+v) + c(u+v) = d\\u^3 + v^3 + (3uv + c)(u+v) = d\\$$
This part is a little strange, from what I've read, this is supposed to be del Ferro's deep insight - it seems a little hacky to me! If we let \(3uv + c = 0\), then \(u^3 + v^3 = d\) and \(v = \frac{-c}{3u}\).
$$u^3 + \big(\frac{-c}{3u}\big)^3 = d \\u^3 - \frac{c^3}{27u^3} - d = 0 \\u^6 - du^3 - \frac{c^3}{27} = 0$$
let \(z = u^3\)
$$z^2 - dz - \frac{c^3}{27} = 0$$
Now using the quadratic formula:
$$z = \frac{d \pm \sqrt{(-d)^2-(4)(1)(\frac{-c^3}{27}})}{2} \\z = \frac{d}{2} \pm \frac{\sqrt{d^2+\frac{4c^3}{27}}}{2} \\z = \frac{d}{2} \pm \frac{\sqrt{d^2+\frac{4c^3}{27}}}{\sqrt{4}} \\z = \frac{d}{2} \pm \sqrt{\frac{d^2}{4}+\frac{4c^3}{(27)(4)}} \\z = \frac{d}{2} \pm \sqrt{\frac{d^2}{4}+\frac{c^3}{27}} \\$$
now substituting back in \(z = u^3\):
$$u^3 = \frac{d}{2} \pm \sqrt{\frac{d^2}{4}+\frac{c^3}{27}} \\u = \sqrt[3]{\frac{d}{2} \pm \sqrt{\frac{d^2}{4}+\frac{c^3}{27}}} \\$$
from above,
$$u^3 + v^3 = d \\v^3 = d-u^3 \\$$
Only consider positive root (using the negative root will result in the same exact answer):
$$v^3 = d - \bigg[\frac{d}{2} + \sqrt{\frac{d^2}{4}+\frac{c^3}{27}}\bigg]\\v^3 = \frac{d}{2} - \sqrt{\frac{d^2}{4}+\frac{c^3}{27}}\\x = u + v \\x = \sqrt[3]{\frac{d}{2} + \sqrt{\frac{d^2}{4}+\frac{c^3}{27}}} + \sqrt[3]{\frac{d}{2} - \sqrt{\frac{d^2}{4}+\frac{c^3}{27}}}$$
We can also obtain another perfectly valid version of the formula by factoring out a \(\sqrt[3]{-1} = -1\) from the second term:
$$ x = \sqrt[3]{\frac{d}{2} + \sqrt{\frac{d^2}{4}+\frac{c^3}{27}}} - \sqrt[3]{-\frac{d}{2} + \sqrt{\frac{d^2}{4}+\frac{c^3}{27}}}$$
This is the formula del Ferro derived. It's crazy to think that ~500 years ago only one person on the planet knew this formula - and he had no idea it would lead to the development and acceptance of imaginary numbers. We end up focusing on a slightly different cubic, introduced later by Cardan:
$$x^3 = cx + d$$
Note that this rearrangement changes the sign of c. Substituting \(-c\) for \(c\) into our final result above, we obtain:
$$x = \sqrt[3]{\frac{d}{2} + \sqrt{\frac{d^2}{4}-\frac{c^3}{27}}} - \sqrt[3]{-\frac{d}{2} + \sqrt{\frac{d^2}{4}-\frac{c^3}{27}}}$$
Just as above, we can factor out a \(\sqrt[3]{-1} = -1\) from the second term to obtain the equivalent formula:
$$x = \sqrt[3]{\frac{d}{2} + \sqrt{\frac{d^2}{4}-\frac{c^3}{27}}} + \sqrt[3]{\frac{d}{2} - \sqrt{\frac{d^2}{4}-\frac{c^3}{27}}}$$
This is version of the cubic formula that will ultimately lead us into issues with \(\sqrt{-1}\), as shown in the video.
WORKS CITED
An Imaginary Tale: The Story of sqrt(-1) - Paul J. Nahin Nahin's book has proven incredibly helpful through the process of creating this series, I highly recommend it for those looking to go deeper into imaginary numbers.
Mathematics and its history - John Stillwell
