What is the degree of the cubic polynomial
Solving cubic equations
Lecture on the mathematical seminar
in the Network Computing course
Technological University Bergakademie Freiburg
Faculty of Mathematics and Computer Science
Institute for Theoretical Mathematics
Prof. Dr. Udo Hebisch
The formula for solving cubic equations was named after Girolamo Cardano (1501-1576). Cardano studied medicine and philosophy in Padua and was already rector of the University of Padua before he was the third attempt to become a Dr. doctorate in medicine.
The mathematician Dal Ferro had found a solution, but did not publish it, but rather communicated it to his students Annibale dalla Nave and Antonio Maria Fior. The latter challenged Tartaglia (approx. 1500-1557), whose real name was Nicolo Fontana, to a mathematical competition in which he deposited 30 tasks with a notary that Tartaglia was supposed to solve. The tasks that Fior had set were all of the type x³ + px = q. Eight days before the end of the competition, Tartaglia found a formula and was able to solve the 30 tasks within 2 hours.
Girolamo Cardano also heard that Tartaglia was in possession of the solution formula and asked him to give him the formula so that he could publish it in his next book under Tartaglia’s name. Tartaglia refused at first, but later handed the formula to Cardano. He promised to keep the formula encrypted. Cardano himself received the formula as a poem .
Quando chel cubo con le cose appresso
If the cube with the couches next to it
A detailed derivation of the formula from the poem can be found on the WWW at http://www.mathematik.uni-kl.de/~meyer/Cardano/card.html.
A cubic equation or equation of the 3rd degree is an equation in which the highest power in which the unknown x occurs is just 3 (i.e. x³):
Solving these equations is much more difficult than solving the quadratic equations, since square roots and cube roots are required to solve them.
The solution of the cubic equation takes place in several steps . First of all, it's best to use the entire equation to be multiplied in order to avoid breaks in the further course of the solution. This gives the following equation:
from which the 2nd and 3rd power are now isolated:
1st step - cubic addition
In this step, similar to the quadratic addition to the quadratic equation, a cubic addition determined. So the left side of the equation can be used as to be written.
According to the binomial formula:
thus also applies:
It follows that the term as cubic addition can be added on both sides.
Then the left side is written as a cube and the right side is summarized
Now the term becomes still excluded. This results in:
This term is now still as written:
2nd step - substitution of x
In the equation:
the following substitution is now carried out:
This results in a new equation of the form:
Now all the terms of the last equation are brought to the left. The result is saved as a reduced cubic equation in the shape:
The reduced cubic equation contains no square link more, however, a linear term 3py, so the equation for not by means of one Cube root can be solved.
3rd step - Representation of the solution by means of two cube roots
The following applies to the reduced cubic equation:
Assuming that y can be represented by the sum of two cube roots u and v, in the reduced cubic equation
Using the binomial formulas we get:
The coefficient comparison method states that this equation is safely met if the following conditions are true:
The first condition can be transformed according to:
The system of equations thus holds for u and v:
4th step - solving the system of equations
The above system of equations is not linear, but nevertheless easy to solve. This is where Vieta's theorem for quadratic equations helps, which says:
Are x1 and x2 Zeroing the quadratic equation:
Applied to the system of equations from step 3 by adding and are set, results in the sum of and equal –q. The product of and is equal to . According to Vieta's theorem, are and the two solutions and the square resolvent:
The solutions of square resolvent but read:
In order to determine u and v, the cube root is now taken from the above term. To avoid a cube root in the denominator, the entire term is expanded with 4.
This results in the following solutions for u and v:
This gives the first solution to the reduced cubic equation:
Through the Back substitution results in a solution of the general cubic equation:
Is the discriminant the quadratic resolvent positive or zero, u and v have real values.
The consideration of real radicands of the cube root can be omitted, since these can always be solved in the real. In this case, therefore, y1 and x1 real solutions of the (reduced) cubic equation.
If the discriminant is negative, the "Casus irreducibilis" one that will be discussed later.
From the Fundamental theorem of algebra (Gauss around 1800) it is known that every algebraic equation of the nth degree has exactly n real or complex solutions. It is also known that every equation of the nth degree has at least one real solution for odd n.
This means that the cubic equation has 3 solutions, one of which is already known. To determine the other two solutions, one applies the polynomial division and splits the appropriate linear factor from the cubic polynomial.
The remainder of the division is zero because y1 satisfies the reduced cubic equation.
The other two solutions y2, y3 of the cubic equation now result from the quadratic equation:
which can be written with uv = -p as follows:
The discriminant of the quadratic equation is:
The discriminant is therefore negative for all real u and v (except u = v). It follows that for the first solution y1 is real, but the two other solutions of the cubic equation are complex conjugate.
The three solutions to the reduced cubic equation are then:
So for u = v another (double) real zero.
This gives the solutions to the general cubic equation:
Treatment of the "casus irreducibilis"
In this case, the quadratic resolvent has a negative discriminant, so that e.g.1 and Z2 become complex. In this case, the following applies:
The two solutions of the quadratic resolvent are conjugate complex. These solutions are now presented in the trigonometric form; H. as a magnitude and argument (phase angle).
According to Moivre's formula, the nth root of a complex number is calculated by taking the root of its absolute value and dividing its argument by n. For the present case, these are the cube root from z1 and Z2:
So u and v can also be represented as follows:
So it's not just z1 and Z2 conjugates complex, but also their cube roots u and v.
Since the sum of two conjugate complex numbers is real (since the imaginary parts cancel each other out due to the different signs), the first solution of the reduced cubic equation is real in this case as well:
The calculation for y3 takes place analogously.
Examples  
A cubic equation
always has three real or one real and two complex zeros.
First, the general cubic equation is replaced by the linear transformation
to the reduced cubic equation
After that becomes the discriminant calculated.
For D> 0 is there a real one and two complex solutions:
In the case D = 0 there are 3 real zeros, at least two of which are equal. The calculation is made as in the case D> 0.
For the case D <0 there are three different real solutions:
The solutions of general cubic equation are in each case:
Solution of cubic equation (PDF)
Further information on the WWW:
 German translation by Prof. Dr. Heinz Lueneburg
 Algebraic equations
 Mathematics Online
 Bartsch - Mathematical formulas
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