Question

x的三次方+px+q=0怎么通解？

Answer 1

x^3 +px + q = 0 的通解是：
x1 = ( -q/2 + ( (q/2)^2 + (p/3)^3 )^(1/2) )^(1/3) + ( -q/2 - ( (q/2)^2 + (p/3)^3 )^(1/2) )^(1/3) ;
x2 = m * ( -q/2 + ( (q/2)^2 + (p/3)^3 )^(1/2) )^(1/3) + m^2 * ( -q/2 - ( (q/2)^2 + (p/3)^3 )^(1/2) )^(1/3) ;
x3 = m^2 * ( -q/2 + ( (q/2)^2 + (p/3)^3 )^(1/镇梁2) )^(1/3) + m * ( -q/2 - ( (q/2)^2 + (p/3)^3 )^(1/御世运2) )^(1/3) ;
其中：m = ( -1 + i * 3^(1/返裤2) )/2 , m^2 = ( -1 - i * 3^(1/2) )/2
而三次方程的一般形式：
ax^3 + bx^2 + cx + d = 0
两边除以a,后设 x = y - b/3a,就可以化成 y^3 + py + q = 0 的形式：
p = c/a - b^2/(3*a^2), q = (2*b^3)/(27*a^3) - (c*b)/(3*a^2) + d/a ;
然后再用上面的公式就行了。其中x1肯定是实根。