Answer
There is a number that is exactly one more than its cube.
Work Step by Step
Suppose that $x$ is exactly one more than its cube:
$x = x^3+1$
$x^3-x+1 = 0$
Let $f(x) = x^3-x+1$
Note that $f$ is a continuous function on $(-\infty, \infty)$.
$f(0) = 1$
$f(-2) = -5$
By the Intermediate Value Theorem, there is a number $c$ in $(-2,0)$ such that $f(c) = 0$. Then $c$ is exactly one more than its cube.
There is a number that is exactly one more than its cube.
