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    I identified that x^2 would become

    But even with me substituing x for z; to give z^3/3-9=0

    I remain unsure how to proceed further.
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    (Original post by apronedsamurai)


    I identified that x^2 would become

    But even with me substituing x for z; to give z^3/3-9=0

    I remain unsure how to proceed further.
    Now solve for Z.
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    (Original post by apronedsamurai)


    I identified that x^2 would become

    But even with me substituing x for z; to give z^3/3-9=0

    I remain unsure how to proceed further.
    Is there an actual question
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    (Original post by uberteknik)
    Now solve for Z.

    I thought it was z^3/3=9

    I then just took an educated guess; that it was 3; because 3^3=27/3=9

    But that was more guesswork and intuition, I am wondering is this like approximiate roots, i.e. you use the trial and error method, or is there a more concrete method/trick to be used
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    (Original post by apronedsamurai)
    I thought it was z^3/3=9

    I then just took an educated guess; that it was 3; because 3^3=27/3=9

    But that was more guesswork and intuition, I am wondering is this like approximiate roots, i.e. you use the trial and error method, or is there a more concrete method/trick to be used
    Why would you need to guess

    Z^3/3 = 9

    Multiplying both sides by 3 gives

    Z^3 = 27

    Cube rooting both sides gives

    Z = 3
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    I wasn't sure that was the approach to take, that is why I am asking....
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    (Original post by apronedsamurai)
    I wasn't sure that was the approach to take, that is why I am asking....
    TenofThem gave you the approach, it's simple manipulation, multiplying, taking cube roots etc etc. You're trying to isolate the z term.
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    (Original post by apronedsamurai)
    I thought it was z^3/3=9

    I then just took an educated guess; that it was 3; because 3^3=27/3=9

    But that was more guesswork and intuition, I am wondering is this like approximiate roots, i.e. you use the trial and error method, or is there a more concrete method/trick to be used
    There's no "guesswork" involved in this - it's just basic GCSE rearrangement.

    If \dfrac{z^3}{3} = 9 then z^3 =  27 and so z = \sqrt[3]{27} = 3
 
 
 
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