Consider be a number exactly and it is greater than its cube.

Therefore .

Then consider the function is .

Intermediate value theorem :

The function is continuous on the closed interval , let be the number between  and , Where then exist a number in such that .

Consider the function is continuous over the interval .

Prove that the number exists between and .

.

Substitute in the above function.

.

Substitute in the above function.

Thus,

The intermediate value theorem says there is exist a root between and .

There is a at least one number exactly and it is greater than its cube.