I won't do this problem, so I'll tell you the general solution of cubic equation with one yuan, which is more practical. Of course, there must be a simple way to solve this problem. ) after changing the highest term coefficient to 1, it is: x 3+ax 2+bx+c = 0.

Let x=y-a/3, and the equation becomes: y 3+py+q = 0.

q=c-ab/3+2a^3/27 p=b-a^2/3

Let y=u+v substitute, and you get: u 3+v 3+3uv (u+v)+p (u+v)+q = 0.

u^3+v^3+q+(u+v)(3uv+p)=0

If: u 3+v 3+q = 0, 3u v+p = 0, and u and v are found, you can get y=u+v as the solution.

u^3+v^3=-q

uv=-p/3，u^3v^3=(-p/3)^3=-p^3/27

V^3 u^3 is the solution of quadratic equation: Z 2+QZ-P 3/27 = 0.

U 3, v 3 = z = (-q √ d)/2, where d = q 2+4p 3/27.

So u and v are: z 1, z2 = 3 √ z.

Let ω=(- 1+i√3)/2, and the three solutions of y are:

y 1=z 1+z2

y2=ωz 1+ω2z2

y3=ω2z 1+ωz2

Therefore:

x 1=y 1-a/3

x2=y2-a/3

x3=y3-a/3

D>0 has a real root and a pair of yoke roots.

D=0 has three real roots, two or three of which are equal.

D<0 has three unequal real roots.

When d < 0, three real roots can be easily found according to a trigonometric identity:

cos 3A=4cos3A-3cosA

z=cosA，z^3-3/4z- 1/4 cos3A=0

Let y=nz and substitute y 3+py+q = 0 to get z 3+z p/n 2+q/n 3 = 0.

Let p/n 2 =-3/4, q/n 3 =- 1/4cos3a, that is.

n=√-4p/3，

cos3A=-4q/n3=-q/2/(√(-p^3/27)

Due to d

The three solutions of Z are: COSA, COS (A+ 120) and COS (A+240).