Solving general cubic without complicated substitutions I am trying to solve a general cubic without complicated substitutions:
$$
ax^3+bx^2+cx +d = 0
$$
Assuming nonzero $a$, after dividing both sides by $a$, moving the $\frac{d}{a}$ term to the RHS and performing further simplifications I get:
$$
(x + \frac{b}{3a}) [ (x+\frac{b}{3a})^2 - \frac{b^2-3ac}{3a^2} ] = \frac{9abc-27a^2d-2b^3}{27a^3}
$$
On the LHS I seem to have a product of a linear function and a quadratic function in a vertex form, however I am stuck as to how to proceed next because of the RHS term.
I notice that the $b^2-3ac$ term is $\Delta_0$ and the $9abc-27a^2d-2b^3$ term is, in my case, $-\Delta_1$ from Wikipedia's general formula of the cubic function.
How can I take it further from here?

Consider an equation of degree $3$ over $\mathbb{C}$: $ R(x)=x^3+ux^2+vx+w=0$ where $u,v,w\in\mathbb{C}$.
Putting $ x=y-u/3$, we obtain an equation in the form $ y^3+px+q=0$.
If $ p=0$, then it's easy, else:
Putting $ y=z+b/z$, we obtain an equation in the form 
(1) $ z^6+a_4z^4+a_3z^3+a_2z^2+a_0=0$.
The system $ a_4=a_2=0$ gives $ b=-p/3$.
With this value of $ b$, (1) is easily solved: $ 2$ values for $ z^3$, therefore $ 6$ values for $ z$. Finally, the formula $ y=z+b/z$ gives $ 2$ times each root of $ y^3+px+q$ and we are done.
Example: $R(x)=x^3-3x^2+x-1$; $x=y+1$ gives $y^3-2y-2=0$. $y=z+2/(3z)$ gives $z^6-2z^3+8/27=0$, that is $z^3=1\pm (1/9)\sqrt{57}$; finally the $6$ values of $z$ are 
$z=\rho (1\pm (1/9)\sqrt{57})^{1/3}$ where $\rho\in\{1,\dfrac{-1\pm i\sqrt{3}}{2}\}$ is a cubic root of $1$.
