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!
 A: Not exactly what you're asking for but the substitution $x=2\sqrt{-p/3}\cos\theta$ turns $x^3+px+q=0$ into $4\cos^3\theta-3\cos\theta-3q=0$. Since $4\cos^3\theta-3\cos\theta=\cos3\theta$, this is $\cos3\theta=3q$, so $\theta=(1/3)\arccos(3q)$, and $x=2\sqrt{-p/3}\cos((1/3)\arccos(3q))$. 
If $p>0$ and you don't want imaginaries then you can start with the analogous "triple angle" formula for the hyperbolic cosine, instead. 
A: Before Viéte, there was Cardano. Suppose you have already depressed the cubic a la Tartaglia, so that it now has the form $x^3+ax+b=0,$ write this as $$x^3=px+q$$ and find $u,v$ satisfying $p=3uv$ and $q=u^3+v^3.$ This then gives you a quadratic whose roots are $u^3,v^3,$ whence you can find $u,v$ and a solution of the original cubic is $x=u+v.$ This gives the well-known formula of Cardano. But the process is easier to remember, obviously.
A: Here is my patent for doing the job. I found it in the school some almost 40 years ago, because i could never remember the "winning substitution" with the target of being prepared for the Olympiads. The idea is to reduce with a linear simple substitution to the equation 
$$
x^3+px+q=0\ .\qquad(*)
$$
Now recall the formula, my favorite formula at those times:
$$
x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\ .
$$
Or even better, but i was not so far, since we need complex numbers and one of the primitive third roots of unity, let us write it like $\varepsilon =\frac 12(-1+\sqrt{-3})$ and delegate the choice of $\sqrt{-3}$,
$$
x^3+y^3+z^3-3xyz=(x+y+z)(x+\varepsilon y+\varepsilon^2 z)(x+\varepsilon^2 y+\varepsilon z)\ .
$$
And we try of course to bring $(*)$ in the above form by searching for some two values $y,z$ such that
$$
\left\{
\begin{aligned}
p &= -3yz\ ,\\
q &= y^3+z^3\ .
\end{aligned}
\right.
$$
From the two equations above, we get immediately the sum and the product of the unknown quantities $y^3$ and $z^3$, this leads (Vieta) to an equation of second degree with these quantities as roots, we get $y$ by a human choice of  the cubic root of $y^3$, then this determines by $p=-3yz$ the correlated cubic root of $z$ to be taken.
Now the roots are not only easily extracted, but their "common structure" is better prepared for a Galois theoretical study, they correspond to the vanishing of the one or the other factors above, so we solve for $x$ in:
$$ 
\begin{aligned}
0 &= x+y+z\ ,&&\text{ or }\\
0 &= x+\varepsilon y+\varepsilon^2 z\ ,&&\text{ or }\\
0 &= x+\varepsilon^2 y+\varepsilon z\ .
\end{aligned}
$$

The above road map is of course not different in the result, although it is different in the exposition, but it was for me differently different, say  psychologically and mnemotechnically, because it was a simple idea behind the needed substitution. And this is still the method i used didactically many times. 
A: 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.
