# Understanding Cubic Equation Formula for Polynomials of Degree Three

• MHB
• gevni
In summary, the conversation is about understanding the parameters of the cubic equation formula for solving polynomial equations of degree three. The person has questions about the "." dot in the formula, whether the formula remains the same for real numbers, and how to use the formula for different types of equations. They also ask about getting three roots from one formula and using it to find the minimum value of x. They are looking for help in understanding the formula or any resources that can help them.
gevni
Hi, Can someone please help me in understanding few parameters of cubic equation formula for solving polynomial of degree three. I attached the formula in the screenshot.
My questions are:
(1) what is ". " dot in the end of the formula and what does it mean?
(2) I want to use it only for real number not complex, so do the formula remain the same?
(3) It states the solution is for this kind of equation$ax^3+ax^2+cx+d=0$ what if I have equation like that $ax^3+ax^2-cx+d=0$? Do i have to put negative value with c parameter in the below formula or whatever the case formula remain same?

(4) It is just one formula then how I can get three roots by using this?
(5) I want to use this formula for getting the minimum possible value of x and if it results in 3 roots then which root show minimum value of x.

I am very thankful if someone help me understanding this formula or any resource that can help me understanding this.

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gevni said:
Hi, Can someone please help me in understanding few parameters of cubic equation formula for solving polynomial of degree three. I attached the formula in the screenshot.
My questions are:
(1) what is ". " dot in the end of the formula and what does it mean?
(2) I want to use it only for real number not complex, so do the formula remain the same?
(3) It states the solution is for this kind of equation$ax^3+ax^2+cx+d=0$ what if I have equation like that $ax^3+ax^2-cx+d=0$? Do i have to put negative value with c parameter in the below formula or whatever the case formula remain same?

(4) It is just one formula then how I can get three roots by using this?
(5) I want to use this formula for getting the minimum possible value of x and if it results in 3 roots then which root show minimum value of x.

I am very thankful if someone help me understanding this formula or any resource that can help me understanding this.

View attachment 9772
The "." means that the sentence is finished. I have been studying the cubic formula for several years. So the formula remain the same. To see what is happening in the formula, it is recommended to derive the formula. Good news, I have a derivation on the cubic formula. I will provide the derivation to you:

Let ##ax^3+bx^2+cx+d=0## where ##a\neq0##

Cbarker1 said:
The "." means that the sentence is finished. I have been studying the cubic formula for several years. So the formula remain the same. To see what is happening in the formula, it is recommended to derive the formula. Good news, I have a derivation on the cubic formula. I will provide the derivation to you:

Let ##ax^3+bx^2+cx+d=0## where ##a\neq0##
Thank you for reply. The part of your post is not displayed which shows the derivation steps. Can you please post them again

I was in the middle of that. However, it was deleted. So sorry for that.

Cbarker1 said:
I was in the middle of that. However, it was deleted. So sorry for that.
No problem but can you please let me know that the same formula that was shown for $ax^3+bx^2+cx+d=0$ we can use for $ax^3+bx^2-cx+d=0$

Cbarker1 said:
The "." means that the sentence is finished. I have been studying the cubic formula for several years. So the formula remain the same. To see what is happening in the formula, it is recommended to derive the formula. Good news, I have a derivation on the cubic formula. I will provide the derivation to you:
Let $ax^3+bx^2+cx+d=0$ where $a\neq0$ and $a,b,c,d$ are arbitrary real numbers. Divide $a$ through the previous equation (since the roots of the previous equation does not change if it is modified by the division): $$x^3+b'x^2+c'x+d'=0 \, \text{where} \, b'=\frac{b}{a}, c'=\frac{c}{a}, \text{and}\,\, d'=\frac{d}{a} \, \text{(1).}$$ By introduction of a new unknown this equation can be simplified, moreover, so that (1) will not have a second power of the unknown. To do what I have described I need to set $x=y+k$ with $k$ still arbitrary. By Taylor's formula (look up in a calculus textbook if you don't know), I will use the first four terms of the formula to determine what the value of $k$: So let $f(x)=x^3+b'x^2+c'x+d'$. $f(y+k)=f(k)+f'(k)y+\frac{f''(k)}{2}y^2+\frac{f'''(k)}{6}y^3$ and $f(k)=k^3+b'k^2+c'k+d'$, $f'(k)=3k^2+2b'k+c'$, $\frac{1}{2}f''(k)=3k+b'$, $\frac{1}{6} f'''(k)=1$. To get rid of the term involving $y^2$, it is suffices to choose $k$ so that $3k+b'=0 \implies k=\frac{-b'}{3}$. We will plugin the value of $k$ into the following functions $f$ and $f'$ and it yields the following values:
$$f\left(\frac{-b'}{3}\right)=d'-\frac{b'c'}{3}+\frac{2{b'}^2}{27} \, \, \text{and} \, \, f'\left(\frac{-b'}{3}\right)=c'-\frac{{b'}^2}{3}.$$ So the substitution for the elimination of $y^2$ is $x=y-\frac{b'}{3}.$ Then the equation (1) is transformed into the following equation through the substitution: $$y^3+py+q=0\, \text{(2)} \, \text{where} \, \, p=c'-\frac{{b'}^2}{3} \, \text{and} \, q= d'-\frac{b'c'}{3}+\frac{2{b'}^2}{27}.$$ A cubic equation of the form (2) can be solved by means of the following device: We seek to satisfy it by setting $y=u+v$, thus introducing two variables $u$ and $v$. On substituting this expression into (2) and arranging terms in a proper way, $u$ and $v$ have to satisfy the equation (let's called it (3): $$u^3+v^3+(p+3uv)(u+v)+q=0,$$ with two variables. This problem is indeterminate unless we find another relationship between $u$ and $v$. This is the relationship that we will be using $3uv+p=0 \implies uv=\frac{-p}{3}$. Then, it follows from (3) that $u^3+v^3=-q$, so that the solution of the cubic (2) can be obtained by solving the system of two equation:$$\begin{cases} u^3+v^3=-q \\ uv=\frac{-p}{3} \end{cases}$$ Taking the cube to the second equation, we will have $u^3v^3=\frac{-p^3}{27}$ and so, from the system of equations and the previous equation, we know the sum and product of the two unknown quantities $u^3$ and $v^3$. (How? There is a formula called the Viete's formula that describes how the roots of a polynomial is related to the coefficient of the same polynomial). These quantities are the roots of the quadratic equation: $t^2+qt-\frac{p^3}{27}=0$. We will separate the roots of the quadratic into $A$ and $B$, respectively: $$A=-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}} \,\text{and}\, B=-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}$$ where we at liberty to the square root as we please. Now owing to the symmetry between the terms $u^3$ and $v^3$ in the systems of equation, we can set $u^3=A$ and $v^3=B$. If some determined value of the cube root of $A$ is denoted $\sqrt[3]{A}$, the three possible values of $u$ will be $u=\sqrt[3]{A}$, $u=\omega\sqrt[3]{A}$, $u={\omega}^2\sqrt[3]{A}$, where $\omega=\frac{-1+i\sqrt{3}}{2}$ is an imaginary cube root of unity. As to $v$, it will have also three values: $v=\sqrt[3]{B}$, $v=\omega\sqrt[3]{B}$, $v={\omega}^2\sqrt[3]{A}$ but not every one of them can be associated with three possible values of $u$, since $u$ and $v$ must satisfy the relation $uv=-\frac{p}{3}$. If $\sqrt[3]{B}$ stands for that cube root of $B$ which satisfies the relation $\sqrt[3]{A}\sqrt[3]{B}=-\frac{p}{3}$, then the values of $v$ that can be associated with $u=\sqrt[3]{A}$, $u=\omega\sqrt[3]{A}$, $u={\omega}^2\sqrt[3]{A}$ will be $v=\sqrt[3]{B}$, $v=\omega\sqrt[3]{B}$, $v={\omega}^2\sqrt[3]{A}$. Hence, equation (2) will have the following roots: $y_1=\sqrt[3]{A}+\sqrt[3]{B}$, $y_2=\omega\sqrt[3]{A}+{\omega}^2\sqrt[3]{B}$, and $y_3=\omega\sqrt[3]{B}+{\omega}^2\sqrt[3]{A}$. After you find what is the value $A$ and $B$, you back-substitute the $y$s into $x=y-\frac{b'}{3}$ in order to fine the value of $x$.

I will post the discussion about the solutions (Discriminant and how to calculate a special case of the sum of two cube roots that leads to an integer answer as well as irreducible case and how to find the roots in that case.) from the book listed below.
Theory of Equation by Uspensky

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This is from Theory of Equations by J.V Upsensky. Copyright in 1948 by Mcgraw-Hill Book Company Inc in New York City. Page 86 starts at the second to the bottom to 87 to the bottom then 88 and so on.

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gevni said:
No problem but can you please let me know that the same formula that was shown for $ax^3+bx^2+cx+d=0$ we can use for $ax^3+bx^2-cx+d=0$
In a word, just replace c with -c in the formula.

-Dan

Cbarker1 said:
Let $ax^3+bx^2+cx+d=0$ where $a\neq0$ and $a,b,c,d$ are arbitrary real numbers. Divide $a$ through the previous equation (since the roots of the previous equation does not change if it is modified by the division): $$x^3+b'x^2+c'x+d'=0 \, \text{where} \, b'=\frac{b}{a}, c'=\frac{c}{a}, \text{and}\,\, d'=\frac{d}{a} \, \text{(1).}$$ By introduction of a new unknown this equation can be simplified, moreover, so that (1) will not have a second power of the unknown. To do what I have described I need to set $x=y+k$ with $k$ still arbitrary. By Taylor's formula (look up in a calculus textbook if you don't know), I will use the first four terms of the formula to determine what the value of $k$: So let $f(x)=x^3+b'x^2+c'x+d'$. $f(y+k)=f(k)+f'(k)y+\frac{f''(k)}{2}y^2+\frac{f'''(k)}{6}y^3$ and $f(k)=k^3+b'k^2+c'k+d'$, $f'(k)=3k^2+2b'k+c'$, $\frac{1}{2}f''(k)=3k+b'$, $\frac{1}{6} f'''(k)=1$. To get rid of the term involving $y^2$, it is suffices to choose $k$ so that $3k+b'=0 \implies k=\frac{-b'}{3}$. We will plugin the value of $k$ into the following functions $f$ and $f'$ and it yields the following values:
$$f\left(\frac{-b'}{3}\right)=d'-\frac{b'c'}{3}+\frac{2{b'}^2}{27} \, \, \text{and} \, \, f'\left(\frac{-b'}{3}\right)=c'-\frac{{b'}^2}{3}.$$ So the substitution for the elimination of $y^2$ is $x=y-\frac{b'}{3}.$ Then the equation (1) is transformed into the following equation through the substitution: $$y^3+py+q=0\, \text{(2)} \, \text{where} \, \, p=c'-\frac{{b'}^2}{3} \, \text{and} \, q= d'-\frac{b'c'}{3}+\frac{2{b'}^2}{27}.$$ A cubic equation of the form (2) can be solved by means of the following device: We seek to satisfy it by setting $y=u+v$, thus introducing two variables $u$ and $v$. On substituting this expression into (2) and arranging terms in a proper way, $u$ and $v$ have to satisfy the equation (let's called it (3): $$u^3+v^3+(p+3uv)(u+v)+q=0,$$ with two variables. This problem is indeterminate unless we find another relationship between $u$ and $v$. This is the relationship that we will be using $3uv+p=0 \implies uv=\frac{-p}{3}$. Then, it follows from (3) that $u^3+v^3=-q$, so that the solution of the cubic (2) can be obtained by solving the system of two equation:$$\begin{cases} u^3+v^3=-q \\ uv=\frac{-p}{3} \end{cases}$$ Taking the cube to the second equation, we will have $u^3v^3=\frac{-p^3}{27}$ and so, from the system of equations and the previous equation, we know the sum and product of the two unknown quantities $u^3$ and $v^3$. (How? There is a formula called the Viete's formula that describes how the roots of a polynomial is related to the coefficient of the same polynomial). These quantities are the roots of the quadratic equation: $t^2+qt-\frac{p^3}{27}=0$. We will separate the roots of the quadratic into $A$ and $B$, respectively: $$A=-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}} \,\text{and}\, B=-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}$$ where we at liberty to the square root as we please. Now owing to the symmetry between the terms $u^3$ and $v^3$ in the systems of equation, we can set $u^3=A$ and $v^3=B$. If some determined value of the cube root of $A$ is denoted $\sqrt[3]{A}$, the three possible values of $u$ will be $u=\sqrt[3]{A}$, $u=\omega\sqrt[3]{A}$, $u={\omega}^2\sqrt[3]{A}$, where $\omega=\frac{-1+i\sqrt{3}}{2}$ is an imaginary cube root of unity. As to $v$, it will have also three values: $v=\sqrt[3]{B}$, $v=\omega\sqrt[3]{B}$, $v={\omega}^2\sqrt[3]{A}$ but not every one of them can be associated with three possible values of $u$, since $u$ and $v$ must satisfy the relation $uv=-\frac{p}{3}$. If $\sqrt[3]{B}$ stands for that cube root of $B$ which satisfies the relation $\sqrt[3]{A}\sqrt[3]{B}=-\frac{p}{3}$, then the values of $v$ that can be associated with $u=\sqrt[3]{A}$, $u=\omega\sqrt[3]{A}$, $u={\omega}^2\sqrt[3]{A}$ will be $v=\sqrt[3]{B}$, $v=\omega\sqrt[3]{B}$, $v={\omega}^2\sqrt[3]{A}$. Hence, equation (2) will have the following roots: $y_1=\sqrt[3]{A}+\sqrt[3]{B}$, $y_2=\omega\sqrt[3]{A}+{\omega}^2\sqrt[3]{B}$, and $y_3=\omega\sqrt[3]{B}+{\omega}^2\sqrt[3]{A}$. After you find what is the value $A$ and $B$, you back-substitute the $y$s into $x=y-\frac{b'}{3}$ in order to fine the value of $x$.

I will post the discussion about the solutions (Discriminant and how to calculate a special case of the sum of two cube roots that leads to an integer answer as well as irreducible case and how to find the roots in that case.) from the book listed below.
Theory of Equation by Uspensky
Thank you for clarifying it.

gevni said:
(2) I want to use it only for real number not complex, so do the formula remain the same?
You can use the same formula, but you probably can't avoid using complex numbers with this formula. If the cubic equation has 3 real roots, the formula requires taking cube root of a complex number. If you want to use only real numbers, you'll need other approach to solve those cubic equations.

(4) It is just one formula then how I can get three roots by using this?

After you've found a root, you can divide the cubic and then use quadratic formula to solve the rest 2 roots.

(5) I want to use this formula for getting the minimum possible value of x and if it results in 3 roots then which root show minimum value of x.

Unfortunately it's not that straightforward. You need to do some kind of analysis beforehand to find out, which method is the most effective. I suggest you to take a look into the English wikipedia. Link here.

## What is a cubic equation?

A cubic equation is a polynomial equation of degree three, meaning the highest exponent in the equation is three. It is written in the form ax^3 + bx^2 + cx + d = 0, where a, b, c, and d are constants and x is the variable.

## What is the cubic equation formula?

The cubic equation formula, also known as the cubic formula or Cardano's formula, is a method for solving cubic equations. It is a complex formula that involves finding the roots of the equation using a combination of addition, subtraction, multiplication, division, and square roots.

## Why is it important to understand the cubic equation formula?

Understanding the cubic equation formula is important because it allows us to solve cubic equations, which are commonly used in many fields of science and engineering. It also helps us to better understand the properties and behavior of cubic equations.

## What are the three types of solutions for cubic equations?

There are three types of solutions for cubic equations: one real root, three real roots, or one real root and two complex roots. The type of solution depends on the discriminant of the equation, which is b^2 - 4ac.

## What are some real-world applications of cubic equations?

Cubic equations have many real-world applications, including in physics, engineering, and economics. For example, they can be used to model the motion of a projectile, calculate the volume of a cube, or determine the optimal price for a product. They are also used in computer graphics to create smooth curves and surfaces.

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