The three roots of a cubic equation? 10

The first root: 1 6*((108*c+12*3 (1 2)*(4*b 3+27*c 2*a) a) (1 2))*a 2) (1 3) a-2*b ((-108*c+12*3 (1 2)*(4*b 3+27*c 2*a) a) (1 2))*a 2) (1 3), the second root: -1 12*((108*c+12*3 (1 2)*(4*b 3+27*c 2*a) a) (1 2))*a 2) (1 3) a+b ((-108*c+12*3 (1 2)*(4*b 3+27*c 2*a) a) (1 2))*a 2) (1 3)+1 2*i*3 (1 2)*(1 6*((108*c+12*3 (1 2)*(4*b 3+27*c 2*a) a) (1 2))*a 2) (1 3) a+2*b ((-108*c+12*3 (1 2)*(4*b 3+27*c 2*a) a) (1 2))*a 2) (1 3)), Third root: -1 12*((108*c+12*3 (1 2)*(4*b 3+27*c 2*a) a) (1 2))*a 2) (1 3) a+b ((-108*c+12*3 (1 2)*(4*b 3+27*c 2*a) a) (1 2))*a 2) (1 3)-1 2*i*3 (1 2)*(1 6*((108*c+12*3 (1 2)*(4*b 3+27*c 2*a) a) (1 2))*a 2) (1 3) a+2*b ((-108*c+12*3 (1 2)*(4*b 3+27*c 2*a) a) (1 2))*a 2) (1 3))

It's complicated, and I came up with a conclusion using a math software, so it's not easy to get a math formula here.

There should be plural numbers in it. --The formula is complicated.

Generally, it is not necessary to directly find the root of the problem of one yuan and three times, but to solve the problem by nature.

x1+x2+x3=-b/a。Veda's theorem describes the relationship between roots and coefficients: the general formula is.

ax 3+bx 2+cx+d=0, and the three roots are x1, x2, x3x1+x2+x3=-b a

x1*x2+x2*x3+x3*x1=c/ax1*x2*x3=-d/a

A standard type of one-dimensional cubic equation.

ax 3+bx 2+cx+d=0(a,b,c,d r, and a≠0), the solution is:

1. Answer: In 1545, the Italian scholar Caldan published the Caldan formula.

Law. 2. The Shengjin formula published by the Chinese scholar Fan Shengjin in 1989.

Law. Both formulas can solve standard one-dimensional cubic equations. It is convenient to use the Caldan formula to solve the problem, in contrast, although the Shengjin formula is simple in form, it is more lengthy and inconvenient to memorize, but the actual solution is more intuitive.

Veda theorem introduces the lead pat relationship between the root and the macro coefficient: the general form of Huai is absolutely enviable.

ax 3+bx 2+cx+d=0, and the three roots are x1, x2, x3x1+x2+x3=-b a

x1*x2+x2*x3+x3*x1=c/ax1*x2*x3=-d/a