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To solve the equation \( a + 3 \) as an extraneous solution, we need to start by solving the equation given:

\[
\frac{x + 6}{x} - \frac{2}{x + 6} = \frac{12}{x^2 + 6}
\]

Let's follow these steps:

1. **Find a common denominator for the left-hand side (LHS):**

   The common denominator for \( \frac{x + 6}{x} \) and \( \frac{2}{x + 6} \) is \( x(x + 6) \):

   \[
   \frac{x + 6}{x} = \frac{(x + 6)^2}{x(x + 6)} = \frac{x^2 + 12x + 36}{x(x + 6)}
   \]
   \[
   \frac{2}{x + 6} = \frac{2x}{x(x + 6)}
   \]

2. **Combine the fractions on the LHS:**

   \[
   \frac{x^2 + 12x + 36}{x(x + 6)} - \frac{2x}{x(x + 6)} = \frac{x^2 + 12x + 36 - 2x}{x(x + 6)} = \frac{x^2 + 10x + 36}{x(x + 6)}
   \]

3. **Equate the LHS to the right-hand side (RHS):**

   \[
   \frac{x^2 + 10x + 36}{x(x + 6)} = \frac{12}{x^2 + 6}
   \]

   Notice that \( x^2 + 6 = (x + \sqrt{6})(x - \sqrt{6}) \), so we write the RHS with a common denominator:

   \[
   \frac{12}{x^2 + 6} = \frac{12}{x(x + 6)}
   \]

4. **Cross-multiply to solve for \( x \):**

   \[
   (x^2 + 10x + 36) \cdot (x^2 + 6) = 12x(x + 6)
   \]

   Simplify the equation by expanding both sides:

   \[
   x^4 + 6x^2 + 10x^3 + 60x + 36x^2 + 216 = 12x^2 + 72x
   \]

   Combine like terms:

   \[
   x^4 + 10x^3 + 42x^2 + 60x + 216 = 12x^2 + 72x
   \]

   Move all terms to one side:

   \[
   x^4 + 10x^3 + 30x^2 - 12x^2 + 60x - 72x + 216 = 0
   \]

   Simplify:

   \[
   x^4 + 10x^3 + 30x^2 - 12x^2 + 60x - 72x + 216 = 0
   \]

   Combine like terms:

   \[
   x^4 + 10x^3 + 18x^2 - 12x + 216 = 0
   \]

5. **Solve for \( x \):**

   To find the extraneous solution, we note that \( a + 3 \) must be a solution to the original equation that does not satisfy the equation. We need to check values for \( x \) where the denominators could be zero.

   If \( x + 3 \) is an extraneous solution, substitute \( x = a + 3 \):

   \[
   \frac{(a + 3) + 6}{a + 3} - \frac{2}{(a + 3) + 6} = \frac{12}{(a + 3)^2 + 6}
   \]

   Simplify the equation and solve for \( a \).

The detailed solution may require finding the roots of the quartic polynomial or using numerical methods if algebraic solution becomes cumbersome. 

Do you want more detailed steps on solving this quartic equation or any specific part of the process?

Here are some relative questions you might find interesting:

1. How do we identify extraneous solutions in general?
2. What are the steps for solving rational equations?
3. How can we simplify complex fractions?
4. Why is it important to check the solutions in the original equation?
5. How do we factorize quartic polynomials?
6. What methods can be used to solve higher-degree polynomials?
7. How do we handle cases where the denominators are zero?
8. What are the common pitfalls in solving equations with rational expressions?

**Tip:** Always verify your solutions in the original equation to ensure they are not extraneous.

Learn how to solve a rational equation involving complex fractions and quartic polynomials step-by-step. This problem covers key concepts such as rational equations, complex fractions, and quartic polynomials. Suitable for high school students.

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