Derivation for the general cubic formula

Question

It's a long equation, and Wikipedia writes it to be $$x_k = -\frac{1}{3a}(b + u_kC + \frac{\Delta_0}{u_kC}), \quad k \in \{1,2,3\}$$

But there is no derivation of it. The sources I've read so far online discuss Cardano's formula and Vieta's substitution, which both rely on depressed cubics. Does this general formula follow the same steps to derive? If so, I've also been wondering why is it that we want to manipulate $x^2$ in $ax^3 + bx^2 + cx + d$ because most papers don't touch on that.

Sorry in advanced if these questions are stupid (recently sustained head injury).

Answer 1

You can start by dividing through by a to get

$$x^3 + (b/a)x^2 + (c/a)x + d/a = 0$$

Then use the substitution $x = u - (b/3a)$ to eliminate the term in $x^2$. This gets you to the reduced form

$$u^3 + Pu + Q = 0$$

Where P and Q can be expressed in terms of a,b,c and d. [exercise for the reader]

Finally use the substitution

$$u = v + R/v$$

Multiplying out the terms in u and $u^3$ we can eliminate $v^2$ and $v^4$ terms if we choose

$$R = -P/3$$

This results in a quadratic in $v^3$

$$v^6 + Qv^3 - P^3 / 27 = 0$$

Solve this by the usual method then reverse through the various substitutions to get back to x.

Answer 2

The Wikipedia article somewhat scrambles the exposition. The first term in $x_k$ is simply the shift used to depress the polynomial (it centers the inflection point). And by looking closely, you will recognize the pattern of Vieta's substitution in the remaining terms.