Algebra/Cubic Equation

In this chapter we will discuss the cubic function in the form $ax^{3}+bx^{2}+cx+d=0$ .

We should note: This subject is much more lengthy and complicated than the quadratic formula, and, oddly enough, includes an inevitable usage of a new mathematical invention called "complex numbers". This subject is not dealt with much in either schools or universities, and we won't be wrong by saying that the average student doesn't know this chapter.

Introduction

About 500 years ago, Italian mathematicians began dealing with this problem. Just like the pretty trivial quadratic algorithm was known, they tried finding a solving algorithm for the cubics.

A few of the mathematicians – Luca Pacioli, Scipione del Ferro, Antonio Fiore, Niccolò Fontana Tartaglia, Gerolamo Cardano, Lodovico Ferrari – had many public competitions. They would keep their methods a secret in order to defeat their opponents in those contests. The ones losing were forced to give up their university jobs for the winners.

Pacioli, perhaps because of a few failed attempts, published a book called Summa de arithmetica in 1494, in which he had claimed it is impossible to solve the cubics algebraically. Some conjectures believe this spurred Del Ferro a few years later to find a solution for all equations of $x^{3}+ax+b=0$ . He kept this a secret for some time until he handed it over to his student Fiore, who used the knowledge while competing against Tartaglia ("Stammerer"). To his astonishment, Tartaglia had found a solution for all equations of $x^{3}+ax^{2}+b=0$ as well as his own solution itself and defeated him.

Now, Tartaglia faced a newer and even tougher rival, Cardano, who after tremendous efforts and persuasion convinced Tartaglia to send him his cubic solutions as a cipher poem, under the promise to keep it a secret, until Tartaglia himself would publish it all in a book. With the help from one of his students, Ferrari, he extended the solution to the larger form of $ax^{3}+bx^{2}+cx+d=0$ , through reducing that form down to the simple $x^{3}+ax+b=0$ . Meanwhile, Ferrari discovered a totally different solution for an even higher degree, the "quartic function".

When Cardano and Ferrari wanted to publish their discoveries in a book of their own, they didn't know how to do so without breaking their promise to Tartaglia. Then, after having a word with Del Ferro's heir, Hannival Nave, they had learned Del Ferro's work predated Tartaglia. This news finally made them break their oath, and they published a book called Ars Magna, to the displeasure of Tartaglia. They quarreled with the furious stammerer, and Tartaglia had finally competed with Ferrari, who defeated him easily.

And what does this have to do with complex numbers?
However, when Cardano extended the solution form, he was astonished to find some equations in which his formula had lead him to an unfathomable expression of negative numbers under a square root. This was an absurdity, since there are no real numbers such that their squaring gives a negative number – adding insult to the injury of not truly understanding the usage of negative numbers themselves.

According to the story it was the equation $x^{3}=15x+4$ which he had tried to factor out its simple result $x=4$ with his algorithm.
The result was:

$x={\sqrt[{3}]{2+{\sqrt {-121}}}}+{\sqrt[{3}]{2-{\sqrt {-121}}}}$

In spite of that result, he understood he must execute operations with this "non-sensible doodling" in front of him, just like negative numbers have a use in finding positive results in quadratics etc.

How did Cardano arrive at the answer above? No need to worry – it will all be clear shortly.

How will we factor out x?

We have the equation $$ax^{3}+bx^{2}+cx+d=0$$ Let's divide it with the coefficient $a\neq 0$ : $$x^{3}+{\frac {b}{a}}x^{2}+{\frac {c}{a}}x+{\frac {d}{a}}=0$$ Completing it to a cube $(x+y)^{3}=x^{3}+3x^{2}y+3xy^{2}+y^{3}$ .

We'll add and subtract a similar cubic expression. By adding and subtracting:

$${\begin{matrix}{\color {Orange}{\Bigg (}}x^{3}+{\color {red}3\left({\dfrac {b}{3a}}\right)x^{2}}+{\color {green}3\left({\dfrac {b}{3a}}\right)^{2}x}+{\color {blue}\left({\dfrac {b}{3a}}\right)^{3}}{\color {Orange}{\Bigg )}}+{\dfrac {c}{a}}x-{\color {green}3\left({\dfrac {b}{3a}}\right)^{2}x}-{\color {blue}\left({\dfrac {b}{3a}}\right)^{3}}+{\dfrac {d}{a}}=0\\\\{\color {Orange}\left(x+{\dfrac {b}{3a}}\right)^{3}}+{\dfrac {c}{a}}x-3\left({\dfrac {b}{3a}}\right)^{2}x-\left({\dfrac {b}{3a}}\right)^{3}+{\dfrac {d}{a}}=0\\\\\left(x+{\dfrac {b}{3a}}\right)^{3}+\left({\dfrac {3ac-b^{2}}{3a^{2}}}\right){\color {JungleGreen}x}+{\dfrac {27a^{2}d-b^{3}}{27a^{3}}}=0\\\\\left(x+{\dfrac {b}{3a}}\right)^{3}+\left({\dfrac {3ac-b^{2}}{3a^{2}}}\right){\color {JungleGreen}\left(x+{\dfrac {b}{3a}}-{\dfrac {b}{3a}}\right)}+{\dfrac {27a^{2}d-b^{3}}{27a^{3}}}=0\\\\\left(x+{\dfrac {b}{3a}}\right)^{3}+\left({\dfrac {3ac-b^{2}}{3a^{2}}}\right){\color {JungleGreen}\left(x+{\dfrac {b}{3a}}\right)}-{\color {JungleGreen}{\dfrac {b}{3a}}}\left({\dfrac {3ac-b^{2}}{3a^{2}}}\right)+{\dfrac {27a^{2}d-b^{3}}{27a^{3}}}=0\\\\\left(x+{\dfrac {b}{3a}}\right)^{3}+\left({\dfrac {3ac-b^{2}}{3a^{2}}}\right)\left(x+{\dfrac {b}{3a}}\right)+{\dfrac {b^{3}-3abc}{9a^{3}}}+{\dfrac {27a^{2}d-b^{3}}{27a^{3}}}=0\\\\{\color {RoyalBlue}\left(x+{\dfrac {b}{3a}}\right)^{3}}+\left({\dfrac {3ac-b^{2}}{3a^{2}}}\right){\color {RoyalBlue}\left(x+{\dfrac {b}{3a}}\right)}+{\dfrac {2b^{3}+27a^{2}d-9abc}{27a^{3}}}=0\\\\{\color {RoyalBlue}y^{3}}+m{\color {RoyalBlue}y}+n=0\end{matrix}}$$

How do we continue?

Surprisingly, this new form is easily solvable, because it fits a cubic form of

$(A\pm B)^{3}\mp 3AB(A\pm B)=A^{3}\pm B^{3}$ A proof of the form here

Let's choose $y=A+B$ without loss of generality. We get

$${\begin{matrix}A+B=y\quad ,\quad -3AB=m\quad ,\quad A^{3}+B^{3}=-n\\\\B=-{\dfrac {m}{3A}}\quad ,\quad B^{3}=-n-A^{3}\\\\A^{3}=-n-\left(-{\dfrac {m}{3A}}\right)^{3}\\\\(A^{3})^{2}+n(A^{3})+\left(-{\dfrac {m}{3}}\right)^{3}=0\\\\A^{3}=-{\dfrac {n}{2}}\pm {\dfrac {1}{2}}{\sqrt {n^{2}-4\left(-{\dfrac {m}{3}}\right)^{3}}}\\\\A={\sqrt[{3}]{-{\dfrac {n}{2}}\pm {\sqrt {\left({\dfrac {n}{2}}\right)^{2}+\left({\dfrac {m}{3}}\right)^{3}}}}}\qquad \qquad B={\sqrt[{3}]{-{\dfrac {n}{2}}\mp {\sqrt {\left({\dfrac {n}{2}}\right)^{2}+\left({\dfrac {m}{3}}\right)^{3}}}}}\\\\x=A+B-{\dfrac {b}{3a}}\end{matrix}}$$

Note

We must consider that over the complex numbers the cubic has 3 solutions, so ${\sqrt[{3}]{1}}$ has 3 roots — 1 real and 2 complex as ${\frac {-1\pm {\sqrt {3}}i}{2}}$ . Therefore When we take the cube root of 1 out of the expressions, we get 3 solutions: $${\color {red}{\begin{matrix}x_{1}={\sqrt[{3}]{-{\dfrac {n}{2}}+{\sqrt {\left({\dfrac {n}{2}}\right)^{2}+\left({\dfrac {m}{3}}\right)^{3}}}}}+{\sqrt[{3}]{-{\dfrac {n}{2}}-{\sqrt {\left({\dfrac {n}{2}}\right)^{2}+\left({\dfrac {m}{3}}\right)^{3}}}}}-{\dfrac {b}{3a}}\\\\x_{2}=\left(-{\dfrac {1-{\sqrt {3}}i}{2}}\right){\sqrt[{3}]{-{\dfrac {n}{2}}+{\sqrt {\left({\dfrac {n}{2}}\right)^{2}+\left({\dfrac {m}{3}}\right)^{3}}}}}+\left(-{\dfrac {1+{\sqrt {3}}i}{2}}\right){\sqrt[{3}]{-{\dfrac {n}{2}}-{\sqrt {\left({\dfrac {n}{2}}\right)^{2}+\left({\dfrac {m}{3}}\right)^{3}}}}}-{\dfrac {b}{3a}}\\\\x_{3}=\left(-{\dfrac {1+{\sqrt {3}}i}{2}}\right){\sqrt[{3}]{-{\dfrac {n}{2}}+{\sqrt {\left({\dfrac {n}{2}}\right)^{2}+\left({\dfrac {m}{3}}\right)^{3}}}}}+\left(-{\dfrac {1-{\sqrt {3}}i}{2}}\right){\sqrt[{3}]{-{\dfrac {n}{2}}-{\sqrt {\left({\dfrac {n}{2}}\right)^{2}+\left({\dfrac {m}{3}}\right)^{3}}}}}-{\dfrac {b}{3a}}\end{matrix}}}$$

Then why do we get complex results as conjugates like this $(a\pm bi)A+(a\mp bi)B$ and not $(a\pm bi)(A+B)$ ?

Simple: Because in one of the first definitions we got $AB=-{\frac {m}{3}}$ , and if we multiply $(a\pm bi)A\cdot (a\mp bi)B$ we'll get $(a^{2}+b^{2})AB$ — moreover the magnitude of these complexes is $\left|{\frac {-1\pm {\sqrt {3}}i}{2}}\right|={\sqrt {a^{2}+b^{2}}}=1$ . See for yourselves.

If we write $(a\pm bi)(A+B)$ we could see this does not solve $AB=-{\frac {m}{3}}$ . Clear and simple.