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

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Discussion Overview

The discussion revolves around the question of whether there exists a number that is exactly one more than its cube. Participants explore this concept through mathematical reasoning and polynomial equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant questions the existence of such a number by stating that if a is the number, then b = a^3 + 1 leads to a contradiction when considering b^3.
  • Another participant proposes that -1.324717957... is a solution to the problem, suggesting that such a number does exist.
  • A further response clarifies the equation a = a^3 + 1, reformulating it as a^3 - a + 1 = 0, and asserts that this polynomial of odd degree must have at least one real solution, indicating the possibility of a real number that meets the criteria.

Areas of Agreement / Disagreement

Participants express differing views on the existence of a number that is one more than its cube, with some asserting it does exist and others challenging that assertion. The discussion remains unresolved.

Contextual Notes

The discussion involves assumptions about polynomial equations and their solutions, as well as the interpretation of the original question regarding the relationship between a number and its cube.

chevyboy86
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is there a number that is exactly one more than its cube?
 
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chevyboy86 said:
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.
 
Yes. That number would be -1.324717957...
 
radou said:
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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