Problem of solving the cubic function

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    Cubic Function
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To solve a cubic function in the form y=ax^3+bx^2+cx+d, plotting the function using software like Maple, MATLAB, or Mathematica is recommended for efficiency. After identifying one root, it can be used as a factor to divide the cubic equation, simplifying it to a quadratic that is easier to solve. For numerical approximations, techniques like Newton's method can provide high-accuracy roots. An alternative approach involves substituting x with z + γ/z to derive a quadratic in z^2, though care must be taken in verifying the solution. Overall, various methods exist, but using software and factoring remains the most straightforward strategy.
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Guys, I may need your help. There is a question saying that how to solve the cubic function in general form, which means that y=ax^3+bx^2+cx+d. How do you guys solve for x? To be honest, I have no idea of this question. Probably, it uses the same way as the quartic function. Thanks!
 
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There are numerous ways of solving cubic functions, but the most efficient way would be to plot it using some type of software--most easily maple, MATLAB, or Mathematica. Once you do that, locate one of the roots. Use the root as a factor, and divide the cubic by that factor to obtain a quadratic. Quadratics are easy to solve, thus you can easily find the remaining two roots.
 
AMenendez said:
There are numerous ways of solving cubic functions, but the most efficient way would be to plot it using some type of software--most easily maple, MATLAB, or Mathematica. Once you do that, locate one of the roots. Use the root as a factor, and divide the cubic by that factor to obtain a quadratic. Quadratics are easy to solve, thus you can easily find the remaining two roots.

Well, if you're going to be content with numerical answers, then there are many good techniques to approximate the roots to a very high degree: http://en.wikipedia.org/wiki/Newton's_method
 
For the cubic equation ax^3+bx^2+cx+d=0 (in your case the constant term is d-y, not d), try substituting x = z +\frac{\gamma}{z}, and solve for z by choosing the constant \gamma correctly. If fairly certain that for a good choice of \gamma (it will become apparent what \gamma must be) you will end up with a quadratic function in z^2.

This way you may arrive at the formula yourself, it's a neat exercise. You probably need to be careful verifying your solution afterwards, as z +\frac{\gamma}{z} is not defined everywhere, and does not attain all values. To make calculations easier, you can assume a = 1 first, and make the necessary modification afterwards.
 
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