Is there a number that is exactly 5 more than its cube?


Question: Is there a number that is exactly 5 more than its cube?

When we multiply a number three times by itself, the resultant number is known as the cube of the original number. 

Answer: Yes

We will prove that there exists a number that is exactly 5 more than its cube by using the Intermediate Value Theorem.

Explanation:

Let x be the number.

x = x3 + 5

Let function f(x) = x3 - x + 5

Since f(-2) = -8+2+5 = -1, f(0) = 5, by intermediate value theorem (IVT) there is at least one zero in (-2, 0).

Therefore there is a number that is exactly 5 more than its cube.