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1. Jinsheng formula

The unary cubic equation ax3+bx2+CX+d = 0, (a, b, c, d∈R, a≠0).

Multiple root discriminant:

a = B2-3ac；

B = 9ad BC

C=c2-3bd，

Total discriminant: δ = B2-4ac.

When A=B=0, Jin Sheng formula ①:

x 1 = X2 = X3 =-b/(3a)=-c/b =-3d/c .

When δ = B2-4ac >; 0, jinsheng formula ②:

x 1 =(-b-(y 1 1/3+y 2 1/3))/(3a)；

X2，3 =(-2 b+ y 1 1/3+y 2 1/3 3 1/2(y 1 1/3-y 2 1/3)I)/(6a)；

Where y 1, 2 = ab+3a (-b (B2-4ac)1/2)/2, I2 =- 1.

When δ= B2-4ac = 0, Formula ③:

x 1 =-b/a+K； X2=X3=-K/2，

Where K=B/A, (A≠0).

When δ = B2-4ac

x 1 =(-b-2a 1/2 cos(θ/3))/(3a)；

X2，3 =(-b+a 1/2(cos(θ/3)3 1/2 sin(θ/3)))/(3a)；

Where θ=arccosT, t = (2ab-3ab)/(2a3/2), (a >；; 0，- 1 & lt； T & lt 1)。

2. Gold filling identification method

①: When A=B=0, the equation has triple real roots;

②: when δ = B2-4ac >; 0, the equation has a real root and a pair of * * * yoke imaginary roots;

③: When δ = B2-4ac = 0, the equation has three real roots, one of which has two roots;

④ When δ = B2-4ac < 0, the equation has three unequal real roots.

3. Golden Sage Theorem

When b=0 and c=0, the formula of "Jin Sheng" is meaningless; When A=0, the golden formula ③ is meaningless; When A≤0, formula ④ is meaningless; When t 1, formula ④ is meaningless.

When b=0 and c=0, does formula ① hold? Is there a value of A≤0 in Formula ③ and Formula ④? Is there a value of t 1 in jinsheng formula ④? The Golden Sage Theorem gives the following answers:

Jinsheng Theorem 1: When A=B=0, if B=0, there must be c=d=0 (at this time, the equation has triple real roots of 0, and Jinsheng formula ① still holds).

Golden Sage Theorem 2: When A=B=0, if b≠0, there must be c≠0 (in this case, use Golden Sage Formula ① to solve the problem).

Golden Sage Theorem 3: When A=B=0, there must be C=0 (at this time, apply Golden Sage Formula ① to solve the problem).

Jinsheng Theorem 4: When A=0, if B≠0, there must be δ > 0 (in this case, Jinsheng Formula ② is used to solve the problem).

Golden Sage Theorem 5: When a < 0, there must be δ > 0 (at this time, apply Golden Sage Formula ② to solve the problem).

Jinsheng Theorem 6: When δ = 0, if B=0, there must be A=0 (at this time, apply Jinsheng formula ① to solve the problem).

Jinsheng Theorem 7: When δ = 0, if B≠0, Jinsheng Formula ③ must have no value of A≤0 (in this case, Jinsheng Formula ③ should be used to solve the problem).

Jinsheng Theorem 8: When δ < 0, Jinsheng Formula ④ must have no value of A≤0. (At this time, apply Jinsheng Formula ④ to solve the problem).

Jinsheng Theorem 9: When δ < 0, T≤- 1 or T≥ 1 in Jinsheng Formula ④ must have no value, that is, the value of T must be-1 < t < 1.

Obviously, when A≤0, there is a corresponding formula to solve the problem.

Note: The inverse of the Golden Sage Theorem may not be true. For example, when δ > 0, there is not necessarily < 0.

The Golden Sage Theorem shows that the Golden Sage formula is always meaningful. Jin Sheng formula can be used to directly solve the univariate cubic equation with arbitrary real coefficients.

When Δ = 0 (d ≠ 0), there is still a prescription to solve the problem by using Caldan formula. Compared with Kadan formula, Jinsheng formula is simpler to express, and it is more intuitive and efficient to use Jinsheng formula to solve problems. Using the discrimination method of gold wealth, the solution of the discrimination equation is intuitive. Multiple root discriminant A = B2-3ac；; B = 9adC = C2-3bd is the simplest formula, and the total discriminant δ = B2-4ac composed of A, B and C is also the simplest formula (it is a very beautiful formula), and its shape is the same as the discriminant of the root of a quadratic equation. The formula (-b (B2-4ac) 1/2)/2 in Jinsheng Formula ② has the form of finding the root of a quadratic equation, which embodies the beauty of order, symmetry, harmony and conciseness in mathematics.

4. Traditional solutions

In addition, the root formula of the unary cubic equation can't be deduced by ordinary deductive thinking, and the standard AX 3+BX 2+CX+D+0 type unary cubic equation can only be formalized into a special type of X 3+PX+Q = 0 by a matching method similar to the root formula of the unary quadratic equation.

The solution of the solution formula of the univariate cubic equation can only be obtained by inductive thinking, that is, the form of the root formula of the univariate cubic equation is summarized according to the form of the root formula of the univariate quadratic equation and the special higher order equation. The formula for finding the root of a univariate cubic equation in the form of x 3+px+q = 0 should be X = A (1/3)+B (1/3), which is the sum of two exponents. Summarized the form of the root formula of the univariate cubic equation, and the next step is to find the content of the square, that is, to represent a and b by P and Q, as follows:

(1) can get the simultaneous cube of X = A (1/3)+B (1/3).

(2)x^3=(a+b)+3(ab)^( 1/3)(a^( 1/3)+b^( 1/3))

(3) Because X = A (1/3)+B (1/3), (2) can be changed to

X 3 = (a+b)+3 (ab) (1/3) x, transpositions are available.

(4) x 3-3 (ab) (1/3) x-(a+b) = 0. Comparing the univariate cubic equation with the special type x 3+px+q = 0,

(5)-3 (AB) (1/3) = P, -(A+B) = Q, simplified.

(6)A+B=-q,AB=-(p/3)^3

(7) In this way, the roots of the univariate cubic equation are formulated into the roots of the univariate quadratic equation, because A and B can be regarded as the two roots of the univariate quadratic equation, and (6) is Vieta's theorem about the two roots of the univariate quadratic equation in the form of ay 2+by+c = 0, namely.

(8)y 1+y2=-(b/a)，y 1*y2=c/a

(9) Comparing (6) and (8), we can make A = Y 1, B = Y2, Q = B/A,-(p/3) 3 = c/a.

(10) because the formula for finding the root of the unary quadratic equation of type ay 2+by+c = 0 is

y 1=-(b+(b^2-4ac)^( 1/2))/(2a)

y2=-(b-(b^2-4ac)^( 1/2))/(2a)

Can become

( 1 1)y 1=-(b/2a)-((b/2a)^2-(c/a))^( 1/2)

y2=-(b/2a)+((b/2a)^2-(c/a))^( 1/2)

Substitute a = y 1, b = y2, q = b/a, -(p/3) 3 = c/a in (9) into (1 1).

( 12)a=-(q/2)-((q/2)^2+(p/3)^3)^( 1/2)

b=-(q/2)+((q/2)^2+(p/3)^3)^( 1/2)

(13) substitute a and b into X = A (1/3)+B (1/3).

( 14)x =(-(q/2)-(q/2)2+(p/3)3)( 1/2))( 1/3)+(-(q/2))

(14) is only the real root solution of a cubic equation with one variable. According to the cubic equation of Vieta's theorem, there should be three roots. But according to the cubic equation of Vieta's theorem, only one root is needed, and the other two roots are easy to find.