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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?

skinny_mike
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  • https://math.stackexchange.com/a/108469/353218 – Darío A. Gutiérrez Feb 12 '18 at 13:09
  • What qualifies as a "complicated substitution" is subjective. For example, is reducing the cubic with the substitution $x = z + \frac{-b}{3a}$, so the sum of the roots $= 0$, complicated or not? If not, then maybe Nickall's geometric analysis of the cubic can provide insights into what to do: http://nickalls.org/dick/papers/maths/cubic1993.pdf – Andy Walls Feb 12 '18 at 15:58
  • I agree that the phrasing is not fortunate, and if somebody finds a better title for this, then I would be glad for them to let me know or edit this question straight away. My point of view is that the very known Cardan's formulae are not a thing (for me), that one can easily come up with. E.g., completing a quadratic function does not require any fancy substitutions to arrive at the solution. Maybe that's just the way it is, it did take some time to solve the cubic in the past. All in all, thank you for a very informative PDF. – skinny_mike Feb 12 '18 at 16:48

1 Answers1

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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 $1$.