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...
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
which gives us
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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