Solution of cubic equation with one variable I want to ask about the root formula ~ don't copy it from the encyclopedia ~ I don't understand ~ read it! I wonder if the cubic equation can be solved only in some special form? What is the root expressed in terms of the corresponding coefficients?

Two methods (I only have high school level)
1: Factorization is written as K * (x-a) (X-B) (x-C) = 0, and then the roots are a, B, C
2: Guess the root, because some can see that there is a root obviously. For example, x ^ 3 + x ^ 2 + x-3 = 0 has a root of 1
Then you can divide (x ^ 3 + x ^ 2 + x-3) by (x-1) = x ^ 2 + 2x + 3
And then it will, right?
Division is like divisor, try it yourself, don't know how to ask me

The solution of cubic equation with one variable
The form of ax ^ 3 + BX ^ 2 + CX + D

The solution is as follows
The formula for finding the root of cubic equation with one variable can not be made by the usual deductive thinking. The matching method similar to the formula for finding the root of quadratic equation with one variable can only formalize the standard cubic equation with one variable of type ax ^ 3 + BX ^ 2 + CX + D = 0 into a special type of x ^ 3 + PX + q = 0
The solution of the formula for solving the cubic equation of one variable can only be obtained by inductive thinking, that is, the form of the formula for finding the root of cubic equation of one variable can be concluded according to the form of the formula for finding the root of linear equation of one variable, quadratic equation of one variable and special higher order equation. The form of the formula for finding the root of cubic equation of one variable, such as x ^ 3 + PX + q = 0, should be x = a ^ (1 / 3) + B ^ (1 / 3), It is the sum of two open cubes. The form of the formula for finding the root of cubic equation with one variable is summed up. The next step is to find out the content in the open cube, that is, to use P and Q to express a and B
(1) Let x = a ^ (1 / 3) + B ^ (1 / 3) be cubic at the same time
（2）x^3=(A+B)+3(AB)^(1/3)(A^(1/3)+B^(1/3))
(3) Because x = a ^ (1 / 3) + B ^ (1 / 3), so (2) can be changed into
X ^ 3 = (a + b) + 3 (AB) ^ (1 / 3) X
(4) Compared with the univariate cubic equation and the special type of x ^ 3 + PX + q = 0, we can see that x ^ 3-3 (AB) ^ (1 / 3) x - (a + b) = 0
(5) - 3 (AB) ^ (1 / 3) = P, - (a + b) = Q
（6）A+B＝－q,AB＝-（p/3）^3
(7) In this way, the root formula of cubic equation with one variable is transformed into the root formula of quadratic equation with one variable, because a and B can be regarded as the two roots of quadratic equation with one variable, and (6) is the Veda theorem about the two roots of quadratic equation with one variable in the form of ay ^ 2 + by + C = 0
（8）y1＋y2＝－（b/a）,y1*y2=c/a
(9) Comparing (6) and (8), we can make a = Y1, B = Y2, q = B / A, - (P / 3) ^ 3 = C / A
(10) Because the root formula of quadratic equation of type ay ^ 2 + by + C = 0 is
y1＝－（b＋（b^2－4ac）^(1/2)）/(2a)
y2＝－（b－（b^2－4ac）^(1/2)）/(2a)
Can be transformed into
(11)y1＝－(b/2a)-((b/2a)^2-(c/a))^(1/2)
y2＝－(b/2a)+((b/2a)^2-(c/a))^(1/2)
Substituting a = Y1, B = Y2, q = B / A, (P / 3) ^ 3 = C / A in (9) into (11), we can get
(12)A＝－(q/2)-((q/2)^2＋（p/3）^3）^(1/2)
B＝－(q/2)+((q/2)^2＋（p/3）^3）^(1/2)
(13) Substituting a and B into x = a ^ (1 / 3) + B ^ (1 / 3), we get
(14)x=（－(q/2)-((q/2)^2＋（p/3）^3）^(1/2)）^(1/3)+（－(q/2)+((q/2)^2＋（p/3）^3）^(1/2)）^(1/3)
Equation (14) is only a real root solution of one variable cubic equation. According to Weida's theorem, one variable cubic equation should have three roots, but according to Weida's theorem, as long as one of the roots is solved, the other two roots can be easily solved
AX3 + bx2 + CX + D = 0 note: P = (27a2d + 9abc-2b3) / (54a3) q = (3ac-b2) / (9A2) X1 = - B / (3a) + (- P + (P2 + Q3) ^ (1 / 2)) ^ (1 / 3) + (- P - (P2 + Q3) ^ (1 / 2)) ^ (1 / 3)

What is the solution of cubic equation of one variable in high school?

The solution of cubic equation of one variable is called grouping method
Generally, there will be quadratic term or primary term in cubic equation. The core idea of grouping method is to group cubic term and secondary term or primary term, then factorize them, and finally form the form of () × () = 0, then make the contents in brackets equal to zero respectively, and then solve them. Generally, the final separated brackets will be a quadratic N-term formula and a primary N-term formula
Example: solve the equation 2x & # 179; + 3x & # 178; = 1
﹙2X³＋2X²﹚＋﹙X²－1﹚＝0
X²﹙X＋1﹚＋﹙X＋1﹚﹙X－1﹚＝0
﹙X＋1﹚﹙X²＋X－1﹚=0
Let (x + 1) and (X & # 178; + x-1) be equal to 0 respectively
The solution is X1 = - 1, X2 = - (1 ＋ 5) / 2, X3 = - (1 ＋ 5) / 2
But this method is not applicable, but most of the cubic equations in high school problems can be solved by this method

Solution of cubic equation with one variable I want to ask about the root formula ~ don't copy it from the encyclopedia ~ I don't understand ~ read it! I wonder if the cubic equation can be solved only in some special form? What is the root expressed in terms of the corresponding coefficients?