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

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The discussion confirms that there exists at least one real number that is exactly one more than its cube. The equation derived from this relationship is a^3 - a + 1 = 0. A specific solution to this equation is approximately -1.324717957. The analysis emphasizes that any polynomial equation of odd degree, such as this one, guarantees at least one real solution.

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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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