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

[chevyboy86]

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

 

[radou]

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.

 

[IMDerek]

Yes. That number would be -1.324717957...

 

[d_leet]

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.