# The solution to a cubic equation

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In summary: RCdtIgxCPOcIn summary, the conversation discusses the solution to the equation ## x^3-x+.1=0 ##, using iterative methods and Vieta's substitution. The equation has 3 real roots, which can be approximated by factoring out ## x-.1 ## and using the quadratic formula. The method for the general solution to the cubic equation is also mentioned, and it is noted that there may be 6 distinct ## w's ## in the solution. The speaker also mentions that they did the calculations by hand, using Taylor series expansions, and that the goal was to show agreement with the quick solution method.
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The solution of ## x^3-x+.1=0 ## is of interest. A quick approximation to the 3 real roots is found. This is compared with the general solution that is found to the equation in the form ## x^3+px=q ##, using the substitution ## x=w-\frac{p}{3w} ##, (the standard way of solving this type of equation), and obtaining a quadratic in ## w^3 ##.
The equation ## x^3-x+.1=0 ##, has 3 real roots that can be quickly approximated as follows: Writing the equation ## x=.1+x^3 ##, iterative methods quickly indicate that there is a root ## r_1 ## near ## x=+.1 ##, and more accurately ## r_1 \approx + .101 ##. ## \\## Doing an approximate calculation of factoring out ## x-.1 ## from ## x^3-x+.1 ## yields ## (x-.1)(x^2+.1x-1 )=0 ## (approximately). ## \\ ## The quadratic formula can be quickly applied to the quadratic factor to give ## r_2 \approx +.95 ## and ## r_3 \approx -1.05 ##. ## \\ ## These results were found to be in agreement with the standard solution to this cubic equation using Vieta's substitution ## x=w-\frac{p}{3w} ##. ## \\ ## Here ## p=-1 ##, and ## q=-.1 ##, for the equation in the form ## x^3+px=q ##. ## \\ ## The substitution gives a quadratic in ## w^3 ## that is ## w^6-qw^3-\frac{p^3}{27}=0 ##. ## \\ ## The number ## w^3 \approx .192 e^{i (\theta +n 360^{\circ}) } ## where ## \theta \approx 105.1^{\circ} ## and ## \theta \approx 254.9^{\circ} ##, and ## n=0,1,2 ##. ## \\ ## The complex number ## w ## is then found as ## w= (.192)^{1/3} e^{i (\theta/3+n 120^{\circ})}=.577 e^{i (\theta/3+n 120^{\circ})} ##. ## \\ ## There are only 3 distinct ## w's ##. ## \\ ## When the computation is then made for ## x ## from ## w ##, the result is ## x=w+\frac{1}{3w}\approx 2(.577) \cos{\phi} ## , and ## x ## is real as it must be. The three roots ## x \approx .101, .95, -1.05 ## are then obtained as above. ## \\ ## Note: I could have kept more decimals in these calculations, but I did them mostly by hand, using Taylor series expansions, etc. to get them reasonably precise. The goal here was simply to show agreement with the very quick solution above.

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chwala, nuuskur and QuantumQuest
One comment on the above: I did this problem mostly for my own interest. I wanted to come up with a cubic equation where iterative methods would supply the first root. The other two roots can then be quickly found (approximately) by factoring out ## x-r_1 ##, and solving the quadratic expression. ## \\ ## I had to google the method for the general solution to the cubic equation. I had seen it in high school, (at least one method of solution), but that was about 45 years ago for me. ## \\ ## One additional comment I have that perhaps someone can give some feedback=I made the statement that there are only 3 distinct ## w's ##: I think there might be 6 distinct ##w's ##. The pair of ##w's ## with ## \phi=##35 degrees ## =105/3 ## and ## 325 ## degrees ##=255/3+240 ## give the same root ## x=+.95 ##, but they are distinct. The quadratic in ## w ^3 ## is a 6th power equation, so it comes as no surprise that there could be 6 different ## w's ##. ## \\ ## Likewise, ## \phi= 105/3+120=155^{\circ} ## and ## 255/3+120=205^{\circ} ## both form the x=-1.05 root, while 105/3+240=275 and 255/3=85 give the x=+.101 root. ## \\ ## I think we can conclude there are 6 distinct ## w's ##.## \\## [Edit: The quadratic equation for ## w^3 ## gets two separate solutions for ## w^3=re^{i \theta} ##, (two different ## \theta's ##), and in subsequently solving for ## w ##, the integer ## n ## above gives 3 distinct values for each ## \theta ##, for ## n=0, 1, 2 ##. (Of course ## n ## can be any integer, but only ## n=0,1,2 ## give distinct values for ## w ##.)]

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I had to google the method for the general solution to the cubic equation. I had seen it in high school, (at least one method of solution), but that was about 45 years ago for
This recent video might by of interest for you:

## 1. What is a cubic equation?

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

## 2. How do you solve a cubic equation?

The most common method for solving a cubic equation is by using the cubic formula. This formula involves finding the roots of the equation using the coefficients a, b, c, and d. Another method is by using factoring, where the equation is rewritten in a factored form and the roots are found from there.

## 3. Are there any special cases when solving a cubic equation?

Yes, there are three special cases that can occur when solving a cubic equation: when the equation has three real roots, when it has one real root and two complex roots, or when it has three equal real roots. These cases can be determined by looking at the discriminant of the equation.

## 4. Can a cubic equation have no real solutions?

Yes, it is possible for a cubic equation to have no real solutions. This occurs when the discriminant of the equation is negative, indicating that the roots are complex numbers. In this case, the equation has two complex roots and one real root.

## 5. How are cubic equations used in real life?

Cubic equations have many practical applications in fields such as engineering, physics, and economics. They can be used to model real-life situations and make predictions about the behavior of systems. For example, they can be used to calculate the trajectory of a projectile or determine the optimal production level for a company.

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