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Write 1729 as the sum of two cubes

  1. Feb 28, 2017 #1
    1. The problem statement, all variables and given/known data
    ##1729## can be written as ##12^3 + 1^3## and ##9^3 + 10^3## and ##7(10 + 9)(12 + 1)##. If
    ##x^3 + (7 - x)^3 = 1729##, use the above to find ##x##. ##x## is a non-integer

    2. Relevant equations
    ##1729 = 12^3 + 1^3 = 9^3 + 10^3 = 7(10 + 9)(12 + 1) = x^3 + (7 - x)^3##

    3. The attempt at a solution
    I used the last equation (the one with x in it) and equated it to 1729 to solve for ##x##. But I need to solve the question without a calculator, and I need to use the information given in the question, which I did not.
    I noticed that the 3rd form equation ## 1729 = 7(10 + 9)(12 + 1)## contains the same terms as those in the cubes, ##(12 + 1)## for ##(12^3 + 1^3)## and ##(10 + 9)## for ##(10^3 + 9^3)##. I think this is important, but I have no idea of how to use this information.
    Thank you for helping
     
  2. jcsd
  3. Feb 28, 2017 #2
    Do you remember how to factor ##(a^3 + b^3)##? Try that and see where it goes.
     
  4. Feb 28, 2017 #3
    I obtain the following:

    ##1729 = 7(9 + 10)(12 + 1) = 7(x^2 - x(7 - x) + (7-x)^2) = (12 + 1)(12^2 - 12 + 1) = (9 + 10)(9^2 - 90 + 10^2)##

    Edit: wrongly wrote a term
     
    Last edited: Mar 1, 2017
  5. Feb 28, 2017 #4

    haruspex

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    I think you mean (7-x)2 as the last term.
     
  6. Mar 1, 2017 #5
    Oh yes. Sorry. Thank you for correcting me
     
  7. Mar 2, 2017 #6

    SammyS

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    Now that you have made that correction, how well are you progressing toward a solution?
     
  8. Mar 2, 2017 #7
    Have you tried what Khashishi suggested in post #2?
     
  9. Mar 2, 2017 #8

    haruspex

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    Isn't that what led to post #3?
     
  10. Mar 3, 2017 #9
    I thought he meant $$(a+b)(a^2-ab+b^2)$$
     
  11. Mar 3, 2017 #10
    I would have set this up differently.
    $$(a+b)[(a+b)^2-3ab]=n[n^2-3x(n-x)]$$where n is a factor of 1729.
     
  12. Mar 3, 2017 #11

    haruspex

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    What is an obvious step from there?
     
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