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

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.

Answer Link

facundo3141592 facundo3141592 · Answer 2 · 2021-09-09T16:11:29+02:00

For a given number X, its cube is written as X^3.

Here we want to see if we can find a number that is exactly 3 more than its cube, this means that:

[tex]X = X^3 + 3[/tex]

We will see that yes, there exists a number with this property.

Then we can try to solve this to find the number with that given property.

We can rewrite it as:

[tex]-X^3 + X - 3 = 0[/tex]

Now we can solve this for X, as the question asks about existence and not for the exact number, we can graph this the above function:

[tex]f(X) = -X^3 + X - 3[/tex]

And if it crosses the x-axis, then at some point we have:

[tex]f(X) = 0[/tex]

This would imply that there exists a real number that is 3 more than its cube.

The graph can be seen below, and we can see that it intersects the X-axis, then yes, there is a number that is exactly 3 more than its cube.

And by watching the graph, we can see that the number is:

[tex]X = -1.672[/tex]

If you want to learn more, you can read:

https://brainly.com/question/18251174