Floor $x$

From ${\color{#3D99F6}x^3} - \lfloor x \rfloor = 3$ . Since the RHS is positive, $x > 0$ . As $\lfloor x \rfloor$ and 3 are integers, $\color{#3D99F6}x^3$ must be an integer. Let $x^3 = n$ . Then $n - \lfloor \sqrt [3]n \rfloor = 3$ , $\implies \lfloor \sqrt [3]n \rfloor = n - 3$ . For $x > 0$ :

$\begin{array} {ccl} \underline{LHS} & \underline{RHS} \\ \lfloor \sqrt [3]n \rfloor = 0 & - 3 < n-3 < - 2 & \small \color{#D61F06} \text{No solution} \\ \lfloor \sqrt [3]n \rfloor = 1 & - 2 < n-3 < 5 & \small \color{#3D99F6} \text{Solution: }n = x^3 = 4 \\ \lfloor \sqrt [3]n \rfloor = 2 & 5 < n-3 < 24 & \small \color{#D61F06} \text{No solution} \end{array}$

It is obvious that for $\lfloor \sqrt [3]n \rfloor \ge 2$ , there is no solution. And $x^3 = \boxed{4}$ is the only solution.