To solve the inequality

\[
\frac{x+6}{(x-8)(x+3)} > 0,
\]

we will follow a structured approach:

### Step 1: Find the critical points
The critical points occur where the expression equals zero or is undefined. These occur when the numerator or the denominator equals zero.

- **Numerator:**
  \[
  x + 6 = 0 \quad \Rightarrow \quad x = -6
  \]
  
- **Denominator:**
  \[
  (x - 8)(x + 3) = 0 \quad \Rightarrow \quad x = 8 \text{ or } x = -3
  \]

Thus, the critical points are \( x = -6, -3, 8 \).

### Step 2: Analyze the intervals
These critical points divide the real line into four intervals:

1. \( (-\infty, -6) \)
2. \( (-6, -3) \)
3. \( (-3, 8) \)
4. \( (8, \infty) \)

We need to test each interval by selecting test points and evaluating the sign of the expression in each interval.

### Step 3: Evaluate the signs
The expression is positive when both the numerator and denominator are either both positive or both negative.

1. **For \( x \in (-\infty, -6) \)**, choose \( x = -7 \):
   \[
   \frac{-7 + 6}{(-7 - 8)(-7 + 3)} = \frac{-1}{(-15)(-4)} = \frac{-1}{60} < 0
   \]

2. **For \( x \in (-6, -3) \)**, choose \( x = -5 \):
   \[
   \frac{-5 + 6}{(-5 - 8)(-5 + 3)} = \frac{1}{(-13)(-2)} = \frac{1}{26} > 0
   \]

3. **For \( x \in (-3, 8) \)**, choose \( x = 0 \):
   \[
   \frac{0 + 6}{(0 - 8)(0 + 3)} = \frac{6}{(-8)(3)} = \frac{6}{-24} = -\frac{1}{4} < 0
   \]

4. **For \( x \in (8, \infty) \)**, choose \( x = 9 \):
   \[
   \frac{9 + 6}{(9 - 8)(9 + 3)} = \frac{15}{(1)(12)} = \frac{15}{12} > 0
   \]

### Step 4: Determine the solution
- The inequality is positive in the intervals \( (-6, -3) \) and \( (8, \infty) \).
- At \( x = -6 \), the numerator is zero, so the expression equals zero (not positive).
- At \( x = -3 \) and \( x = 8 \), the denominator is zero, making the expression undefined.

### Final Answer:
The solution to the inequality is:
\[
x \in (-6, -3) \cup (8, \infty)
\]

Would you like more details or have any questions?

### Related Questions:
1. How do we handle inequalities with more complicated expressions?
2. What happens if both the numerator and denominator are quadratic?
3. Can this method be applied to rational functions with higher powers?
4. How would you solve this inequality if there were absolute values involved?
5. How do you graph the solution set of a rational inequality?

### Tip:
Always check whether the expression is defined at critical points, as division by zero leads to undefined values.

Learn how to solve the rational inequality (x+6)/[(x-8)(x+3)] > 0 with a detailed step-by-step solution. This problem covers algebraic concepts like critical points, sign analysis, and rational expressions. Suitable for high school students in Grades 10-12.

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