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# Solving Cubic equations Watch

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1. I am trying to find a general formula for solving cubics. I tried the most general problem i.e. with ax^3+bx^2+cx+d=0. But I wasn't getting anywhere. So I tried ax^3=cx+d and I seemed to have come up with a rather ugly looking formula that works. Now I need to find a more general solution i.e. introducing the quadratic term again. I have tried everything I can think of and now I am trying to substitute an expression so I can reduce it to the form solved earlier. Can anyone suggest a substitution that might work? Is substitution even the best way to go about solving this problem?
2. It's been done.

Divide through by a to get x^3+Bx^2+Cx+D=0 and then sub in y-B/3 for x. This removes the squared term and then the method you worked out will finish it.
3. (Original post by Get me off the £\?%!^@ computer)
It's been done.

Divide through by a to get x^3+Bx^2+Cx+D=0 and then sub in y-B/3 for x. This removes the squared term and then the method you worked out will finish it.
Really, by who? omg thanks so much I spent hours trying all sorts of weird and different substitutions and didn't even think to divide by a. I think though now the method is more of a procedure than a formula.
4. (Original post by anshul95)
Really, by who?
Tartaglia/Cardano.
5. (Original post by Mr M)
Tartaglia/Cardano.
Thank you - seemed like the Cardano guy had a similar approach as me (yeah). I read some more into it and apparently Tartaglia had some maths "duel" about solving cubics
6. Intriguing!
7. Ever seen this beast of a formula?
8. (Original post by nuodai)
Ever seen this beast of a formula?
OMG I thought the formula I found looked bad. Looking at this one makes wonder how my calculator can solve cubic equations!

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