The problem asks us to solve the equation $\lfloor 2x^3 - x^2 \rceil = 18x - 9$ for $x \in \mathbb{R}$, where $\lfloor x \rceil$ denotes the ceiling function of $x$, defined as the smallest integer greater than or equal to $x$, and $\lfloor x \rfloor$ denotes the floor function of $x$, defined as the largest integer less than or equal to $x$.

1. Problem Description

The problem asks us to solve the equation $\lfloor 2x^3 - x^2 \rceil = 18x - 9$ for $x \in \mathbb{R}$, where $\lfloor x \rceil$ denotes the ceiling function of $x$, defined as the smallest integer greater than or equal to $x$, and $\lfloor x \rfloor$ denotes the floor function of $x$, defined as the largest integer less than or equal to $x$.

2. Solution Steps

The given equation is $\lfloor 2x^3 - x^2 \rceil = 18x - 9$.

Since the ceiling function returns an integer, $18x-9$ must be an integer. Thus, $18x$ is an integer.

Let $18x = n$, where $n$ is an integer. Then, $x = \frac{n}{18}$.

Substitute $x = \frac{n}{18}$ into the equation:

$\lfloor 2(\frac{n}{18})^3 - (\frac{n}{18})^2 \rceil = 18(\frac{n}{18}) - 9$

$\lfloor \frac{2n^3}{18^3} - \frac{n^2}{18^2} \rceil = n - 9$

$\lfloor \frac{2n^3}{5832} - \frac{n^2}{324} \rceil = n - 9$

$\lfloor \frac{n^3}{2916} - \frac{n^2}{324} \rceil = n - 9$

Since $n-9$ is an integer, we have:

$n-9 \le \frac{n^3}{2916} - \frac{n^2}{324} < n-8$

Consider the equation $\frac{n^3}{2916} - \frac{n^2}{324} = n-9$.

$n^3 - 9n^2 = 2916n - 26244$

$n^3 - 9n^2 - 2916n + 26244 = 0$

If $x=1$, $18x-9=9$.

If $x=0.5$, $18x-9 = 0$, which is an integer.

Try $x=0.5 = \frac{9}{18}$. $n=9$.

$\lfloor \frac{9^3}{2916} - \frac{9^2}{324} \rceil = \lfloor \frac{729}{2916} - \frac{81}{324} \rceil = \lfloor \frac{1}{4} - \frac{1}{4} \rceil = \lfloor 0 \rceil = 0$.

$n-9 = 9-9 = 0$. So $x = 0.5$ is a solution.

If $x=2$, $18x-9 = 36-9 = 27$.

Try $x=2$.

$\lfloor 2(2^3) - 2^2 \rceil = \lfloor 2(8) - 4 \rceil = \lfloor 16 - 4 \rceil = \lfloor 12 \rceil = 12$.

$18(2) - 9 = 36 - 9 = 27$.

So $x=2$ is not a solution.

If $n = 54$, then $x=3$.

$\lfloor 2(3^3) - 3^2 \rceil = \lfloor 2(27) - 9 \rceil = \lfloor 54 - 9 \rceil = \lfloor 45 \rceil = 45$.

$18(3) - 9 = 54 - 9 = 45$.

So $x=3$ is a solution.

If we test integer values for $n$, we can try to find integer roots for the equation $n^3 - 9n^2 - 2916n + 26244 = 0$.

Since $x=0.5$, we have $n=9$.

Since $x=3$, we have $n=54$.

Thus, $(n-9)$ and $(n-54)$ are factors.

$(n-9)(n-54) = n^2 - 63n + 486$

$n^3 - 9n^2 - 2916n + 26244 = (n-9)(n^2 +an +b) = n^3 +an^2 + bn - 9n^2 - 9an - 9b = n^3 + (a-9)n^2 + (b-9a)n - 9b$.

$a-9 = -9$ implies $a=0$.

$b-9a = -2916$ implies $b = -2916$.

$-9b = 26244$ implies $b = -2916$.

Therefore, $(n-9)(n^2 - 2916) = (n-9)(n-54)(n+54)=0$.

So $n=9, n=54, n=-54$.

$x = \frac{9}{18} = 0.5$.

$x = \frac{54}{18} = 3$.

$x = \frac{-54}{18} = -3$.

Check $x=-3$.

$\lfloor 2(-3)^3 - (-3)^2 \rceil = \lfloor 2(-27) - 9 \rceil = \lfloor -54 - 9 \rceil = \lfloor -63 \rceil = -63$.

$18(-3) - 9 = -54 - 9 = -63$.

So $x = -3$ is also a solution.

3. Final Answer

The solutions are $x = -3$, $x = 0.5$, and $x = 3$.