<s>[INST] Solving $ax^3+bx^2+cx+d=0$ using a substitution different from Vieta's? We all know, a general cubic equation is of the form

$$ax^3+bx^2+cx+d=0$$ where  $$a\neq0.$$

It can be easily solved with the following simple substitutions:

$$x\longmapsto x-\frac{b}{3a}$$

We get,

$$x^3+px+q=0$$ where, $p=\frac{3ac-b^2}{3a^2}$ and $q=\frac{2b^3-9abc+27a^2d}{27a^3}$ 

Then, using the Vieta substitution,

$$x\longmapsto x-\frac{p}{3x}$$

We get,

$$(x^3)^2-q(x^3)-\frac1{27}p^3=0$$ 

which is can easily turn into a quadratic equation, using the substitution: $x^3 \longmapsto x.$
And here is my question:

In mathematics is there a substitution that is "different" from the  substitution  $x\longmapsto x-\frac{p}{3x}$ that can be used for the standard form cubic equation $x^3+px+q=0$ , which is can easily turn into a quadratic equation?

I'm curious, if there's a new substitute I don't know about.
Thank you!
 [/INST] Possible substitution $x\to X+\frac{1}{z}$, where $X$ is some parameter and $z$ is new unknown, but additional need use some properties of cubic polynomial, which historically we received from Lagrange.
Thus we have solution of equation $f=ax+bx^2+cx^3$:
$\begin{cases}
X=\dfrac{-(a b + 9 c f) + \sqrt{(a b + 9 c f)^2-4 (b^2 - 3 a c) (a^2 + 3 b f)}}{2 (b^2 - 3 a c)}\\
C=a X + b X^2 + c X^3 - f\\
B_1=a + 2 b X + 3 c X^2\\
W=B_1^3-27cC^2\\
B_2=\left\{W^{1/3}\,,-(-1)^{1/3} W^{1/3}\,,(-1)^{2/3}W^{1/3}\right\}\\
x=X+\frac{3C}{B_2-B_1}
\end{cases}$
Deriving this solution see in Intro, and there is a code to verifing in Wolfram and Geogebra.
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