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Is there a number that is exactly one more than its cube?

  1. Jan 19, 2007 #1
    is there a number that is exactly one more than its cube?
     
  2. jcsd
  3. Jan 19, 2007 #2

    radou

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    Take some number a. Its cube is a^3. Let b = a^3 + 1. But b^3 = (a^3 + 1)^3. So, no, it doesn't.
     
  4. Jan 19, 2007 #3
    Yes. That number would be -1.324717957...
     
  5. Jan 19, 2007 #4
    Why are you finding b^3? If we want the number to be a then the equation that says a is one more than its cube is

    a=a3+1
    which gives us
    a3-a+1=0

    And if we note that any polynomial equation of odd degree has at least one solution in the reals then we are assured that this equation has a solution. So there is at least one real number that is one more than its cube.
     
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