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    For a given equation
    ax^3+cx=d , to find x , it says use u-v=d, uv=(c/3)^3, and then x=\sqrt[3]{u}-\sqrt[3]{v}}.

    My question is how do I find this underlined part , :confused: ??( the book does not mention it)
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    Solve
    u-v=d
    and
    uv=(c/3)^3
    simultaneously.

    (For a given equation you will know c and d.)
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    (Original post by rnd)
    Solve
    u-v=d
    and
    uv=(c/3)^3
    simultaneously.

    (For a given equation you will know c and d.)
    My question is how do I deirive [ u-v=d , and uv = (c/3)^3,x=\sqrt[3]{u}-\sqrt[3]{v}}}.


    given only that ax^3+cx=d
    soory if my question isnt specific
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    Are you sure you want to derive those equations?

    If you just want to solve them then you have u=d+v, sub that in uv=(c/3)^3 rarrange to get a quadratic equation and solve it to get v in terms of c and d.

    But how can you not be using a?
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    yes, I want to know how they found the equations(I think the bok assumes its obvious :rolleyes: .I know how to find x using u and v.
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    (Original post by rbnphlp)
    yes, I want to know how they found the equations(I think the bok assumes its obvious :rolleyes: .I know how to find x using u and v.
    OK. I'm giving up on this one then. Hope someone else helps you out.

    My point about a still stands though. How can x depend on c and d but not a? It can't. What's the book by the way?
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    It looks like the standard cubic solution equation in the special case where b=0 and a=1. There's a derivation of the solution on this page:

    http://en.wikipedia.org/wiki/Cubic_e...cubic_function
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    (Original post by rnd)
    OK. I'm giving up on this one then. Hope someone else helps you out.

    My point about a still stands though. How can x depend on c and d but not a? It can't. What's the book by the way?
    History of mathematics quite intresting
 
 
 
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Updated: July 15, 2009

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