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To find the power series for \( f(x) = \frac{6}{(x+1)^3} \), we use the given series and differentiate:

### Step-by-Step Solution

1. **Given Series:**
   \[
   \frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n, \quad |x| < 1
   \]

2. **Function Representation:**
   \[
   f(x) = \frac{6}{(x+1)^3} = 6 \cdot \frac{d^2}{dx^2} \left( \frac{1}{x+1} \right)
   \]

3. **First Derivative:**
   \[
   \frac{d}{dx} \left( \frac{1}{1+x} \right) = \frac{d}{dx} \left( \sum_{n=0}^{\infty} (-1)^n x^n \right) = \sum_{n=1}^{\infty} (-1)^n n x^{n-1}
   \]

4. **Second Derivative:**
   \[
   \frac{d^2}{dx^2} \left( \frac{1}{1+x} \right) = \sum_{n=2}^{\infty} (-1)^n n(n-1) x^{n-2}
   \]

5. **Multiply by 6:**
   \[
   f(x) = 6 \cdot \sum_{n=2}^{\infty} (-1)^n n(n-1) x^{n-2}
   \]

6. **Rewrite the Series:**
   \[
   f(x) = \sum_{n=0}^{\infty} (-1)^{n+2} (n+2)(n+1) x^n
   \]

7. **Interval of Convergence:**
   The interval of convergence remains \((-1, 1)\) due to differentiation not affecting the radius.

### Final Answers

- **Power Series:**
  \[
  f(x) = \sum_{n=0}^{\infty} (-1)^{n+2} (n+2)(n+1) x^n
  \]

- **Interval of Convergence:**
  \((-1, 1)\)

Would you like more details or have any questions?

Here are some questions you might ask:
1. How is differentiation applied to power series?
2. Why does the interval of convergence remain the same?
3. What are common applications of series expansion?
4. How can we verify the power series result?
5. What happens when we differentiate power series multiple times?

**Tip:** When differentiating a power series, adjust indices and factorial terms accordingly.

Explore the power series expansion for the function f(x) = 6/(x+1)^3 with a detailed step-by-step solution. This problem covers power series, differentiation of series, and interval of convergence concepts. Suitable for advanced undergraduate students and beyond.

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