Is there a number that is exactly 3 more than its cube?

Question

Mathias Hancock

Mathematics

18 November, 06:40

Is there a number that is exactly 3 more than its cube?

+3

Answers (1)

Know the Answer?

Jocelyn Rose · Answer 1

Yes, there are three of them. Two of them are complex.

They are the solutions to (x-x^3=3).

The exact value for the real solution is:

(sqrt (239) / (2*3^ (3/2)) - 3/2) ^ (1/3) + 1 / (3 * (sqrt (239) / (2*3^ (3/2)) - 3/2) ^ (1/3))

Approximately, it is x=-1.67169988 ...

If you are expecting integers (which was not specified), then there is none.