Not OP, but I derived it 2 days ago while using toilet. 

And it was for the most basic version of the problem:

\( x^3 + x + a = 0 \)

But it can probably be transformed into any cubic formula if you set \(x = sy+t \) and multiply left side by some constant. 

I didn't really derive it, I just guessed it should be in form:


\( x = \sqrt[3]{p + q} + \sqrt[3]{p - q} \)

When I use it in the equation, I get:

\( (\sqrt[3]{p + q} + \sqrt[3]{p - q})^3 + \sqrt[3]{p + q} + \sqrt[3]{p - q} + a = 0 \)

\( p + q + 3\sqrt[3]{p + q}\sqrt[3]{p + q}\sqrt[3]{p - q} + 3\sqrt[3]{p + q}\sqrt[3]{p - q}\sqrt[3]{p - q} + p - q  \) 
\( + \sqrt[3]{p + q} + \sqrt[3]{p - q} + a = 0 \)

\( 2p + 3\sqrt[3]{p + q}\sqrt[3]{p^2 - q^2} + 3\sqrt[3]{p - q}\sqrt[3]{p^2 - q^2} + \sqrt[3]{p + q} + \sqrt[3]{p - q} + a = 0 \)

\( 2p + 3(\sqrt[3]{p + q} + \sqrt[3]{p - q})\sqrt[3]{p^2 - q^2} + \sqrt[3]{p + q} + \sqrt[3]{p - q} + a = 0 \)

\( (2p + a) + ( 1 + 3\sqrt[3]{p^2 - q^2})(\sqrt[3]{p + q} + \sqrt[3]{p - q}) = 0 \)

In order for this equation to be satisfied I'll break this equation in two:

\(
\begin{cases}

2p + a = 0 \\

( 1 + 3\sqrt[3]{p^2 - q^2})(\sqrt[3]{p + q} + \sqrt[3]{p - q}) = 0

\end{cases}
\)

\(
\begin{cases}

p = -\frac{a}{2} \\

1 + 3\sqrt[3]{p^2 - q^2} = 0

\end{cases}
\)

Then we solve \(q\):

\(1 + 3\sqrt[3]{-\frac{a^2}{4} - q^2} = 0\)

\(\sqrt[3]{-\frac{a^2}{4} - q^2} = -\frac{1}{3}\)

\(-\frac{a^2}{4} - q^2 = -\frac{1}{27}\)

\(q^2 = -\frac{a^2}{4} + \frac{1}{27}\)

\(q = \sqrt{-\frac{a^2}{4} + \frac{1}{27}}\)

and thus:

\( x = \sqrt[3]{-\frac{a}{2} + \sqrt{-\frac{a^2}{4} + \frac{1}{27}}} + \sqrt[3]{-\frac{a}{2} - \sqrt{-\frac{a^2}{4} + \frac{1}{27}}} \)


Also shoutout to the admin. Some fan of Lukyon on soyjak.sharty thinks that you are the good person to discover if manga Lukyon is a virgin. A bit silly question, but he really begged me to ask it. :(