scott wrote:
>> Unless my eyes deceive me, this is equivilent:
>>
>> J = 2a^3 - 9ab + 26c
>> K = Sqrt(J^2 + 4(-a^2 + 3b)^3)
>> M = (1 + i Sqrt(3))/2
>> N = (-1 + i Sqrt(3))/2
>>
>> r1 = -(a/3) +   Cbrt((-J + K)/54) +   Cbrt((-J - K)/54)
>> r2 = -(a/3) + M Cbrt((-J + K)/54) + N Cbrt((-J - K)/54)
>> r3 = -(a/3) + N Cbrt((-J + K)/54) + M Cbrt((-J - K)/54)
>>
>> While hardly on the same level as the quadratic, it's not "that" hard.
> 
> Did you try to derive those equations from ax^3+bx^2+cx+d=0 ?? :-)

No - because it solves the *monic* equation x^3 + ax^2 + bx + c = 0. :-P 
(Notice the absence of a "d" in any of the equations above.)

I'm surprised nobody has yet pointed out that it's "+ 27c" and not "+ 26c".

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