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.