When solving equations, algebraists hope to theoretically prove the existence of solutions and find methods to obtain them. This led to important theories such as the root formulas for equations of degrees 1 to 4 and the proof that algebraic equations of degree 5 and above have no general solution in radicals. However, typical scholars (such as physicists and astronomers) often do not require these theoretical depths; they only care about how to find the roots of a specific equation (especially real roots) as quickly as possible. Therefore, they usually focus on numerical algorithms for equations. Of course, having a relatively simple root formula is also something they desire. The content to be introduced next is a root formula for cubic equations that meets this need, where a significant part of the theory used is related to trigonometric functions.
Prerequisites
\frac{2}{\tan 2A} = \frac{1}{\tan A} - \tan A \frac{2}{\sin 2A} = \frac{1}{\tan A} + \tan A \cos(3A) = 4\cos^3 A - 3\cos A
Approach
Here we only discuss the solution of the cubic equation x^3 + ax^2 + bx = c, as this form is more representative in practical applications. Any such cubic equation can be transformed into the form y^3 + my = n through a substitution, and then further into the form z^3 \pm 3z = p. Since z^3 \pm 3z = p has a relatively simple solution, the roots of the cubic equation can be solved relatively easily (Remember, "relatively" is emphasized here, because no matter how simple the solution for a cubic equation is, it cannot be simpler than the quadratic formula...).
Transformation
I. x^3 + ax^2 + bx = c \to y^3 + my = n
This is actually very simple. Just let x = y - a/3 and substitute it into the original equation to get: y^3 + \left(b - \frac{a^2}{3}\right)y = c + \frac{ab}{3} - \frac{2a^3}{27} Comparing this with y^3 + my = n, it is easy to see that: m = b - \frac{a^2}{3}, \quad n = c + \frac{ab}{3} - \frac{2a^3}{27} The transformation is complete.
II. y^3 + my = n \to z^3 \pm 3z = p
The "\pm" here has the same sign as m. Let y = \lambda z, substitute and rearrange to get z^3 + \frac{m}{\lambda^2}z = \frac{n}{\lambda^3}. Comparing this with z^3 \pm 3z = p, we get: \begin{aligned} \frac{m}{\lambda^2} = \pm 3 &, \quad \frac{n}{\lambda^3} = p \\ \lambda = |m/3|^{1/2} &, \quad p = n |3/m|^{3/2} \end{aligned} The transformation is complete.
Solution
Now the problem is concentrated on solving equations of the type z^3 \pm 3z = p. We need to discuss this in cases.
First, consider z^3 + 3z = p. Let z = A^{-1} - A. Substituting this into the equation and rearranging yields A^{-3} - A^3 = p. This can be turned into a quadratic equation by substitution. However, we will not do that. Instead, let A^3 = \tan B. Using formula (1) from the "Prerequisites," we get \frac{2}{\tan 2B} = \frac{1}{\tan B} - \tan B = p, which means \tan 2B = \frac{2}{p}. Solving backwards, we get A = \left(\tan \frac{\arctan \frac{2}{p}}{2}\right)^{1/3}. This completes the solution. As seen from the graph, for all p, this equation has one and only one real root.
Next, consider z^3 - 3z = p. Let z = A^{-1} + A. Substituting this into the equation and rearranging yields A^{-3} + A^3 = p. Similarly, this can be turned into a quadratic equation. Following the previous method, let A^3 = \tan B. Using formula (2) from the "Prerequisites," we get \frac{2}{\sin 2B} = \frac{1}{\tan B} + \tan B = p, which means \sin 2B = \frac{2}{p}. Solving backwards, we get A = \left(\tan \frac{\arcsin \frac{2}{p}}{2}\right)^{1/3}.
Some friends might think the solution is complete. However, we notice that the expression for A contains the term \arcsin \frac{2}{p}, which only works for |p| > 2. But as seen from the graph, when |p| \leq 2, the equation does not lack real roots; rather, it has 2 to 3 real roots. Thus, we need another method to solve this special case. This time, we use formula (3) from the prerequisites. The equation z^3 - 3z = p can be rewritten as 4(z/2)^3 - 3(z/2) = p/2. The coefficients of each term now correspond to formula (3). Thus, we have \cos 3A = p/2 and 2\cos A = z. Solving backwards gives z = 2\cos\left(\frac{\arccos(p/2) + 2k\pi}{3}\right) for k=0, 1, 2. This provides the three real roots. At this point, the solution for the cubic equation is basically complete.
Summary
The above process is summarized here:
For a general cubic equation x^3 + ax^2 + bx = c, first perform the following calculations: m = b - \frac{a^2}{3}, \quad n = c + \frac{ab}{3} - \frac{2a^3}{27}, \quad p = n \left| \frac{3}{m} \right|^{3/2}
If m is positive or zero, it indicates the equation has only 1 real root; proceed to step (I). If m is negative, then: if |p| > 2, the equation has 1 real root; if |p| = 2, it has 2 real roots; if |p| < 2, it has 3 real roots; proceed to step (II). If m=0, the equation has 1 real root; proceed to step (III).
(I). (m > 0) \begin{aligned} A = \left(\tan \frac{\arctan \frac{2}{p}}{2}\right)^{1/3}, \quad z = A^{-1} - A \\ x = |m/3|^{1/2}z - a/3 \end{aligned}
(II).1. (m < 0, |p| > 2) \begin{aligned} A = \left(\tan \frac{\arcsin \frac{2}{p}}{2}\right)^{1/3}, \quad z = A^{-1} + A \\ x = |m/3|^{1/2}z - a/3 \end{aligned}
(II).2. (m < 0, |p| < 2) \begin{aligned} z = 2\cos\left(\frac{\arccos(p/2) + 2k\pi}{3}\right), \quad (k=0, 1, 2) \\ x = |m/3|^{1/2}z - a/3 \end{aligned}
(III). (m = 0) x = \sqrt[3]{n} - \frac{a}{3}
Conclusion
Through the above transformations and solutions, we can see that any cubic equation with real coefficients can be transformed into the form z^3 \pm 3z = p (where p is real), and z^3 \pm 3z = p always has at least one real root. This means we have also proven that any cubic equation has no fewer than one real root.
One might ask, what if I want to find the non-real roots? This can be done using Vieta’s formulas. For the three roots of the cubic equation x^3 + ax^2 + bx = c, the following relationships hold: x_1 x_2 x_3 = c; \quad x_1 + x_2 + x_3 = -a If one real root is already known, the above equations become a system of two equations with two variables, which can be easily solved. Details are omitted here.
The idea of combining trigonometric functions with the roots of cubic equations has existed for a long time, but it had not been generalized. By studying "Introduction to Celestial Mechanics," I gained inspiration, and through the internet, I found relevant formulas. By organizing these gathered materials, I arrived at this set of solutions for cubic equation roots. Since none of the formulas used were "originally created" by me, I do not dare to arrogantly claim to have "invented" a new method. The inspiration simply came from a persistence in solving equations and the help of the internet and books, so it is hardly worth mentioning. However, I still hope that when readers read or repost this, they can provide a link to Scientific Space (kexue.fm), so that more people can learn about it ^_^.
When reposting, please include the original address: https://kexue.fm/archives/831
For more details on reposting, please refer to: Scientific Space FAQ