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We are given the limit:

\[
\lim_{{x \to -1}} \left( \frac{1}{x + 1} - \frac{3}{x^3 + 1} \right)
\]

### Step-by-step solution:

1. **Check if direct substitution works:**
   First, substitute \( x = -1 \) into the expression:

   \[
   \frac{1}{-1 + 1} - \frac{3}{(-1)^3 + 1} = \frac{1}{0} - \frac{3}{-1 + 1} = \frac{1}{0} - \frac{3}{0}
   \]

   Since this leads to a division by zero (undefined), we cannot directly substitute. This suggests we should look for another method, such as factoring or using limits.

2. **Factor the denominators:**
   Let's focus on the second fraction \( \frac{3}{x^3 + 1} \). We can factor \( x^3 + 1 \) using the sum of cubes formula:
   \[
   x^3 + 1 = (x + 1)(x^2 - x + 1)
   \]
   So, the expression becomes:
   \[
   \frac{1}{x + 1} - \frac{3}{(x + 1)(x^2 - x + 1)}
   \]

3. **Combine into a single fraction:**
   Now we can combine the two fractions:
   \[
   \frac{1}{

lim (1 / (x + 1) - 3 / (x^3 + 1)) as x approaches -1

This problem involves finding the limit of a rational expression by factoring using the sum of cubes formula. The detailed solution covers algebraic manipulation and common limit strategies. Suitable for advanced high school students (Grades 11-12).

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