This note has been used to help create the Solving Cubic Equations - Lagrange's Resolvent wiki

We all can solve quadratic equations of the form $ax^2 + bx + c = 0$ where $a \neq 0$. Simple, isn't it. Even, we have also devised formulaes like:

$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

for that general quadratic. Has anyone wondered about solving higher degree equations? Cubics (of degree 3), Quartics (of degree 4), Quintics (of degree 5), sextics ( of degree 6), etc.? Probably. We might have heard about Cardano's formula in Cubics, Ferrari's solution in quartics. Even, we might know about the Tschirnhausen Transformation and all to obtain depressed equations to solve the original. Hey Wait, we also have a solution for cubics using trigonometry.

We'll discuss here a way to solve cubics. One is the Traditional Cardano's Solution as we all know. Let's explore the rest.

• Lagrange's Resolvents ( A Brief Idea ) :

Guys, do you know about the Abel-Ruffini Theorem? Maybe. Forget that. I'm presenting you a statement, and if you don't know it, just accept it for the present moment.

Galois theory has established that formulas using a finite number of arithmetic operations and root extractions are impossible for general equations of degree greater than four.

What? Really? So, can't we generalize formulas for degree 5+ equations?

Joseph-Louis Lagrange (1736-1813) had something in store. What Lagrange realized was that to solve equations of prime degree $n$ with rational coefficients, one has to solve a resolvent equation of degree $n-1$ also with rational coefficients, which are now called Lagrange resolvents. Please remember that he is talking about prime degrees, like cubics, quintics, heptics, degrees - 11, 13, and so on.

Now here comes our Mathematics!

Using Lagrange's resolvents, to solve the cubic, one has to first solve a quadratic.

Given the general cubic, $x^3 + ax^2 + bx + c = 0$

it's resolvent equation is given by

$z^2 + (2a^3 - 9ab + 27c)z + (a^2-3b)^3 = 0$

such that the solution to the cubic is $x= \dfrac{-a + z_1^{1/3} + z_2^{1/3}}{3}$ where $z_1, z_2$ are roots of the resolvent.

One can note two things. First, the solution is analogous to the quadratic formula. And while Galois theory has established that formulas using a finite number of arithmetic operations and root extractions are impossible for general equations of degree greater than four, there are particular equations solvable as such. So the solution to the solvable quintic should be similar in form, namely,

$x = \dfrac{-a + z_1^{1/5} + z_2^{1/5} + z_3^{1/5} + z_4^{1/5} }{5}$

where $z_i$ are the roots of the Lagrange's quartic resolvent.

We can obtain similar solutions for solvable degrees - 7, 11, 13, 17, ..

Let's take an example:

Let the cubic be $x^3 - 5x^2 + x - 7 = 0$

Resolvent is $z^2 -394z + 22^3 = 0$

with roots $z_i = 197 \pm \sqrt{3129}$

Thus $x = \large{\boxed{ \dfrac{5 + \left(197 + \sqrt{3129} \right)^{1/3} + \left(197 - \sqrt{3129} \right)^{1/3}}{3} }} \approx 5.07475$

And we know how to obtain the rest two roots of a cubic if we are presented with one of the roots.

The important issue here is to find the Lagrange's Resolvents for other higher degrees. Can you think of some?

In the subsequent issues, I'll be posting various other ways to solve a cubic like - Cardano's solution, Translation, Linear Fractional Transformations. Note by Satyajit Mohanty
4 years, 2 months ago

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At last I will get rid of factorization and long division method.

- 4 years, 2 months ago

No.. Factorization and Division Method will be easy for cubic equations having simple rational/integral roots. My note is targeted towards those who are interested in finding the exact roots of any cubic given.

- 4 years, 2 months ago

Awesome! I previously thought that while solving cubic equations one root is to be guessed

- 4 years, 2 months ago

Nah! Well, this method is only one of the ways. I'll post various more ways in the future :)

- 4 years, 2 months ago

Waiting for those posts! :)

- 3 years, 11 months ago

Ya, me too. He must be prepping for entrance exams nowadays.

- 3 years, 11 months ago

Thank you !

- 4 years, 2 months ago

But do you think that this can be very useful? Obviously it is somewhat useful but while solving cubic equations we can use Vieta's or equation is framed such that it's one root is can be guessed easily . This is not very useful until question is asked directly on it . What do you think about that??

- 4 years, 2 months ago

Look at the example I took:

$x^3 - 5x^2 + x - 7 = 0$

Try your Vieta's ? :D

- 4 years, 2 months ago

Yah but knowing all methods is good

- 4 years, 2 months ago