Solve the inequality `x[x]-x^(3)-3[x]+3xgt0` where [.] denote greatest integer function.

Answer

To solve the inequality \( x[\text{x}] - x^3 - 3[\text{x}] + 3x > 0 \), where \([\text{x}]\) denotes the greatest integer function, we can follow these steps:

### Step 1: Rewrite the inequality
We start by rewriting the inequality:
\[ x[\text{x}] - x^3 - 3[\text{x}] + 3x > 0 \]
This can be rearranged as:
\[ x[\text{x}] + 3x - 3[\text{x}] - x^3 > 0 \]

### Step 2: Factor out common terms
We can group the terms involving \([\text{x}]\):
\[ [\text{x}](x - 3) + 3x - x^3 > 0 \]

### Step 3: Isolate \([\text{x}]\)
Now we isolate \([\text{x}]\):
\[ [\text{x}] > \frac{x^3 - 3x}{x - 3} \]

### Step 4: Define the function
Let’s define the function:
\[ f(x) = \frac{x^3 - 3x}{x - 3} \]
This function is undefined at \(x = 3\).

### Step 5: Find the critical points
To find the critical points of \(f(x)\), we set the numerator equal to zero:
\[ x^3 - 3x = 0 \implies x(x^2 - 3) = 0 \]
This gives us the critical points:
\[ x = 0, \quad x = \sqrt{3}, \quad x = -\sqrt{3} \]

### Step 6: Analyze the intervals
We need to analyze the sign of \(f(x)\) in the intervals determined by the critical points:
1. \( (-\infty, -\sqrt{3}) \)
2. \( (-\sqrt{3}, 0) \)
3. \( (0, \sqrt{3}) \)
4. \( (\sqrt{3}, 3) \)
5. \( (3, \infty) \)

### Step 7: Test points in each interval
- For \(x < -\sqrt{3}\), choose \(x = -2\):
\[ f(-2) = \frac{(-2)^3 - 3(-2)}{-2 - 3} = \frac{-8 + 6}{-5} = \frac{-2}{-5} > 0 \]
- For \(-\sqrt{3} < x < 0\), choose \(x = -1\):
\[ f(-1) = \frac{(-1)^3 - 3(-1)}{-1 - 3} = \frac{-1 + 3}{-4} = \frac{2}{-4} < 0 \]
- For \(0 < x < \sqrt{3}\), choose \(x = 1\):
\[ f(1) = \frac{1^3 - 3(1)}{1 - 3} = \frac{1 - 3}{-2} = \frac{-2}{-2} > 0 \]
- For \(\sqrt{3} < x < 3\), choose \(x = 2\):
\[ f(2) = \frac{2^3 - 3(2)}{2 - 3} = \frac{8 - 6}{-1} = \frac{2}{-1} < 0 \]
- For \(x > 3\), choose \(x = 4\):
\[ f(4) = \frac{4^3 - 3(4)}{4 - 3} = \frac{64 - 12}{1} = 52 > 0 \]

### Step 8: Determine intervals where \(f(x) < [\text{x}]\)
From the analysis:
- \( (-\infty, -\sqrt{3}) \): \(f(x) > 0\)
- \((- \sqrt{3}, 0)\): \(f(x) < 0\)
- \((0, \sqrt{3})\): \(f(x) > 0\)
- \((\sqrt{3}, 3)\): \(f(x) < 0\)
- \((3, \infty)\): \(f(x) > 0\)

### Step 9: Combine results
The inequality \( [\text{x}] > f(x) \) holds in the intervals:
- \( (-\infty, -\sqrt{3}) \)
- \( (0, \sqrt{3}) \)
- \( (3, \infty) \)

### Final Solution
Thus, the solution to the inequality \( x[\text{x}] - x^3 - 3[\text{x}] + 3x > 0 \) is:
\[ x \in (-\infty, -\sqrt{3}) \cup (0, \sqrt{3}) \cup (3, \infty) \]