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Math Help - Confused

  1. #1
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    Confused

    number 8 is written as 3 components sum like that, where the first 2 cube amount(a^3+b^3.that means this sum is the lowest), plus 3rd component (C), multiplied by 9, is the lowest.

    It is also known, that the first component is 2 times greater than second one. Find those components.

    8 = a + b + c

    (a^3 + b^3 + c) * 9 > 0

    2a = b

    I do get 81a^3 - 27a +64 = 0 and I believe this is nonse, what am I doing wrong? ;\

    The answer is 2; 1; 5.

    Thank you.
    Last edited by Ellla; January 26th 2011 at 11:45 AM.
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  2. #2
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    It's should be: a=2b (It is also known, that the first component is 2 times greater than second one)

    I don't understand this line:


    "Number 8 is written as 3 component's sum, when the first two compenents third degree sum, plus 3rd component, multiplied by 9,### is the lowest###."
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  3. #3
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    Quote Originally Posted by Ellla View Post
    Number 8 is written as 3 component's sum, where the first two compenents third degree sum, plus 3rd component, multiplied by 9, is the lowest.

    It is also known, that the first component is 2 times greater than second one. Find those components.

    8 = a + b + c

    (a^3 + b^3 + c) * 9 > 0

    2a = b

    I do get 81a^3 - 27a +64 = 0 and I believe this is nonse, what am I doing wrong? ;\

    The answer is 2; 1; 5.

    Thank you.
    If a, b, c are positive integers

    a+b+c=8

    \left(a^3+b^3+c\right)9=minimum.

    a=2b\Rightarrow\ 3b+c=8\Rightarrow\ c=8-3b

    \left[(2b)^3+b^3+8-3b\right]9=min

    \left(8b^3+b^3+8-3b\right]9=min

    81b^3-27b+72=min

    b=1 will minimize this, from which you can find a, c.
    Last edited by Archie Meade; January 26th 2011 at 12:04 PM. Reason: typo
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  4. #4
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    Well, number 8 is written as 3 components sum like that, where the first 2 cube amount(a^3+b^3.that means this sum is the lowest), plus 3rd component (C), multiplied by 9, is the lowest.
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  5. #5
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    Sorry, I might have confused you ;\

    Archie Meade, Im very thankful for your work, but would you mind telling me how do I solve that 81a^3 - 3b + 8 = 0 ?
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  6. #6
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    Quote Originally Posted by Ellla View Post
    Sorry, I might have confused you ;\

    Archie Meade, Im very thankful for your work, but would you mind telling me how do I solve that 81b^3 - 3b + 8 = 0 ?
    I felt that "lowest" may have meant "smallest possible" from the available integers.

    If "b" is not an integer, then you have to use a root-finding method for cubic polynomials.

    If "b" is a positive integer, and we are looking for a minimum, then the 81b^3 term dominates,
    hence by inspection "b" is 1,
    since higher values of "b" will give far greater results.
    I think that's the context of the question, otherwise "b" would be a tiny fraction.
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  7. #7
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    Alright, thanks.
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