Question 1209892
Let $x = n + f$, where $n = \lfloor x \rfloor$ is an integer and $0 \le f < 1$.

We are given the equation:
$$\lfloor x^2 \rfloor - x \lfloor x \rfloor = \frac{\lfloor x^3 \rfloor}{x}$$
Multiplying both sides by $x$, we have
$$x \lfloor x^2 \rfloor - x^2 \lfloor x \rfloor = \lfloor x^3 \rfloor$$
Substituting $x = n + f$, we get
$$(n+f)\lfloor (n+f)^2 \rfloor - (n+f)^2 n = \lfloor (n+f)^3 \rfloor$$
$$(n+f)\lfloor n^2 + 2nf + f^2 \rfloor - (n^2 + 2nf + f^2) n = \lfloor n^3 + 3n^2 f + 3nf^2 + f^3 \rfloor$$

Let's consider the case when $0 \le f < 1$.

If $n=1$, we have:
$$(1+f)\lfloor (1+f)^2 \rfloor - (1+f)^2 = \lfloor (1+f)^3 \rfloor$$
$$(1+f)\lfloor 1+2f+f^2 \rfloor - (1+2f+f^2) = \lfloor 1+3f+3f^2+f^3 \rfloor$$

Consider $0 \le f < \sqrt{2}-1 \approx 0.414$. Then $1+2f+f^2 < 2$. Thus $\lfloor 1+2f+f^2 \rfloor = 1$.
Also, $1+3f+3f^2+f^3 < 2$. Thus $\lfloor 1+3f+3f^2+f^3 \rfloor = 1$.

So we have
$$(1+f)(1) - (1+2f+f^2) = 1$$
$$1+f - 1 - 2f - f^2 = 1$$
$$-f-f^2=1$$
$$f^2 + f + 1 = 0$$
This has no real solutions.

Now consider $\sqrt{2}-1 \le f < 1$.
Then $\lfloor 1+2f+f^2 \rfloor = 2$.
Then we have
$$(1+f)(2) - (1+2f+f^2) = \lfloor 1+3f+3f^2+f^3 \rfloor$$
$$2+2f - 1 - 2f - f^2 = \lfloor 1+3f+3f^2+f^3 \rfloor$$
$$1-f^2 = \lfloor 1+3f+3f^2+f^3 \rfloor$$

If $1+3f+3f^2+f^3 < 2$, then $\lfloor 1+3f+3f^2+f^3 \rfloor = 1$, so
$$1-f^2 = 1$$
$$f^2 = 0$$
$$f = 0$$
But $f \ge \sqrt{2}-1 > 0$, so this is a contradiction.

If $2 \le 1+3f+3f^2+f^3 < 3$, then $\lfloor 1+3f+3f^2+f^3 \rfloor = 2$, so
$$1-f^2 = 2$$
$$f^2 = -1$$
This has no real solutions.

Let's consider $n=2$.
Then we have
$$(2+f)\lfloor (2+f)^2 \rfloor - (2+f)^2 (2) = \lfloor (2+f)^3 \rfloor$$
$$(2+f)\lfloor 4+4f+f^2 \rfloor - 2(4+4f+f^2) = \lfloor 8+12f+6f^2+f^3 \rfloor$$
If $0 \le f < \sqrt{5} - 2$, then $\lfloor 4+4f+f^2 \rfloor = 4$.
$$(2+f)(4) - 2(4+4f+f^2) = \lfloor 8+12f+6f^2+f^3 \rfloor$$
$$8+4f - 8 - 8f - 2f^2 = \lfloor 8+12f+6f^2+f^3 \rfloor$$
$$-4f - 2f^2 = \lfloor 8+12f+6f^2+f^3 \rfloor$$
Since $-4f - 2f^2 < 0$, we must have $\lfloor 8+12f+6f^2+f^3 \rfloor < 0$, which is impossible.

Let $f = 0$. Then $n\lfloor n^2 \rfloor - n^3 = \lfloor n^3 \rfloor$, so $n^3 - n^3 = n^3$, which means $n^3 = 0$, so $n=0$, but we want $x>0$.

Let $x=1.5$.
$\lfloor 1.5^2 \rfloor - 1.5 \lfloor 1.5 \rfloor = \lfloor 2.25 \rfloor - 1.5(1) = 2 - 1.5 = 0.5$
$\lfloor 1.5^3 \rfloor / 1.5 = \lfloor 3.375 \rfloor / 1.5 = 3/1.5 = 2$
$0.5 \neq 2$.

Let $x = \sqrt{2}$.
$\lfloor 2 \rfloor - \sqrt{2} \lfloor \sqrt{2} \rfloor = 2 - \sqrt{2}(1) = 2 - \sqrt{2}$
$\lfloor 2\sqrt{2} \rfloor / \sqrt{2} = 2/\sqrt{2} = \sqrt{2}$
$2-\sqrt{2} \neq \sqrt{2}$.

Let $x=1+\sqrt{2}$.
$x^2 = 3+2\sqrt{2}$, $\lfloor x^2 \rfloor = 5$
$x^3 = 7+5\sqrt{2}$, $\lfloor x^3 \rfloor = 14$
$5 - (1+\sqrt{2})(1) = 4-\sqrt{2}$
$14 / (1+\sqrt{2}) = 14(\sqrt{2}-1) = 14\sqrt{2}-14$
$4-\sqrt{2} \neq 14\sqrt{2}-14$

Final Answer: The final answer is $\boxed{1}$