一元三次方程的解法及求根公式

解：
原方程整理得：
5x^3-60x^2+384x-768=0
(此时，x不等于-4且x不等于4)
经过解方程可知：
5x^3-60x^2+384x-768=0 的三个根为：
实数根：
x1=-4/5*(25+20*10^(1/2))^(1/3)+12/(25+20*10^(1/2))^(1/3)+4
共轭复数根分别为：
x2=2/5*(25+20*10^(1/2))^(1/3)-6/(25+20*10^(1/2))^(1/3)+4+2*i*3^(1/2)*(-1/5*(25+20*10^(1/2))^(1/3)-3/(25+20*10^(1/2))^(1/3))

x3=2/5*(25+20*10^(1/2))^(1/芹竖3)-6/(25+20*10^(1/2))^(1/3)+4-2*i*3^(1/2)*(-1/巧友5*(25+20*10^(1/2))^(1/3)-3/(25+20*10^(1/2))^(1/3))

精确到小数点后14位得：
X1=3.13368638035959
X2=4.43315680982021+5.41875211332654*i
X3=4.43315680982021-5.41875211332654*i

分别将x1,x2,x3带入y=(48-12*x+x^2)/(4+x)-x^2/(16-4*x)检验得：
x1带入y=-4.4409e-015 接近于0
x2带入y=4.4409e-016 +6.6613e-016i 接近于0+0i=0
x3带入y= 4.4409e-016 -6.6613e-016i 近于0+0i=0
所以误差小于10的-15次方
a*x^3+b*x^2+c*x+d=0（a不为0）
这个三次方程的求根公式为：
x1= 1/6/a*(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)-2/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)-1/3*b/a

x2= -1/12/a*(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/嫌宽大2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)+1/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)-1/3*b/a+1/2*i*3^(1/2)*(1/6/a*(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)+2/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3))

x3=-1/12/a*(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)+1/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)-1/3*b/a-1/2*i*3^(1/2)*(1/6/a*(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3)+2/3*(3*c*a-b^2)/a/(36*c*b*a-108*d*a^2-8*b^3+12*3^(1/2)*(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)^(1/2)*a)^(1/3))
这个公式有点长，都不敢写了，怕你吓坏，想了想还是提出来吧。
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