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

We will prove that there exists a number that is exactly 5 more than its cube by using the Intermediate Value Theorem. The IVT states that if a function f is continuous on an interval [a,b] and if f(a) < 0 and f(b) > 0 (or vice-versa), then there is a third point c with a < c < b so that f(c) = 0.

## Explanation:

Let x be the number.

x = x^{3} + 5

Let function f(x) = x^{3} - x + 5

Since f(-2) = -8 + 2 + 5 = -1 < 0, f(0) = 5 > 0, and the function f(x) is continuous in the interval [-2, 0], using intermediate value theorem (IVT) there is at least one point c in (-2, 0) such that f(c) = 0.

⇒ c^{3} - c + 5 = 0

⇒ c = c^{3} + 5