Decide what values of the variable cannot possibly be solutions for the equation. Do not solve. $$ \frac{1}{x-4}+\frac{1}{x+2}=\frac{1}{x^{2}-2 x-8} $$ What values of $$ x $$ cannot be solutions of the equation? $$ \square $$ (Use a comma to separate answers as needed.)

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To determine the values of $$ x $$ that cannot be solutions for the equation

$$
\frac{1}{x-4}+\frac{1}{x+2}=\frac{1}{x^{2}-2 x-8},
$$

we need to identify the values that would make any denominator in the equation equal to zero, as division by zero is undefined.

1. **Identify the denominators:**
   - The left side has two denominators: $$ x - 4 $$ and $$ x + 2 $$.
   - The right side has the denominator $$ x^2 - 2x - 8 $$.

2. **Set each denominator to zero and solve for $$ x $$:**
   - For $$ x - 4 = 0 $$:
     $$
     x = 4
     $$
   - For $$ x + 2 = 0 $$:
     $$
     x = -2
     $$
   - For $$ x^2 - 2x - 8 = 0 $$, we can factor it:
     $$
     x^2 - 2x - 8 = (x - 4)(x + 2) = 0
     $$
     This gives us the same solutions:
     $$
     x = 4 \quad \text{and} \quad x = -2
     $$

3. **Conclusion:**
   The values of $$ x $$ that cannot be solutions of the equation are $$ 4 $$ and $$ -2 $$.

Thus, the final answer is:

$$ 4, -2 $$
$$ x = 4, -2 $$

Decide what values of the variable cannot possibly be solutions for the equation. Do not solve. \( \frac{1}{x-4}+\frac{1}{x+2}=\frac{1}{x^{2}-2 x-8} \) What values of \( x \) cannot be solutions of the equation? \( \square \) (Use a comma to separate answers as needed.)

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