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

  • Thread starter chevyboy86
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  • #1
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is there a number that is exactly one more than its cube?
 

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  • #2
radou
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is there a number that is exactly one more than its cube?

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.
 
  • #3
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Yes. That number would be -1.324717957...
 
  • #4
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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.

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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