To expand \(\left(\frac{1 - x}{1 + x}\right)^n\) up to the \(x^3\) term, we will use the **binomial series expansion** for both the numerator and denominator and then combine them.

### Step-by-Step Expansion

#### 1. **Rewrite the expression:**
We can express the fraction \(\frac{1 - x}{1 + x}\) as:
\[
\left(\frac{1 - x}{1 + x}\right)^n = (1 - x)^n \cdot (1 + x)^{-n}
\]

#### 2. **Expand \((1 - x)^n\) using the binomial theorem:**
For \((1 - x)^n\), the binomial expansion up to \(x^3\) term is:
\[
(1 - x)^n = 1 - nx + \frac{n(n-1)}{2}x^2 - \frac{n(n-1)(n-2)}{6}x^3 + \cdots
\]

#### 3. **Expand \((1 + x)^{-n}\) using the binomial theorem:**
For \((1 + x)^{-n}\), the binomial expansion up to \(x^3\) term is:
\[
(1 + x)^{-n} = 1 - nx + \frac{n(n+1)}{2}x^2 - \frac{n(n+1)(n+2)}{6}x^3 + \cdots
\]

#### 4. **Multiply the two expansions:**
Now multiply the two expansions together, keeping terms only up to \(x^3\):
\[
(1 - nx + \frac{n(n-1)}{2}x^2 - \frac{n(n-1)(n-2)}{6}x^3) \cdot (1 - nx + \frac{n(n+1)}{2}x^2 - \frac{n(n+1)(n+2)}{6}x^3)
\]

#### 5. **Collect terms up to \(x^3\):**
Let's collect the terms up to \(x^3\) by multiplying and adding similar powers of \(x\):

- **Constant term:** \(1 \times 1 = 1\)

- **Linear term (in \(x\)):**
\[
- nx \times 1 + 1 \times (-nx) = -2nx
\]

- **Quadratic term (in \(x^2\)):**
\[
\frac{n(n-1)}{2}x^2 \times 1 + (-nx) \times (-nx) + 1 \times \frac{n(n+1)}{2}x^2 = \frac{n(n-1)}{2}x^2 + n^2x^2 + \frac{n(n+1)}{2}x^2 = n(2n-1)x^2
\]

- **Cubic term (in \(x^3\)):**
\[
-\frac{n(n-1)(n-2)}{6}x^3 \times 1 + \frac{n(n-1)}{2}x^2 \times (-nx) + (-nx) \times \frac{n(n+1)}{2}x^2 + 1 \times -\frac{n(n+1)(n+2)}{6}x^3
\]
This simplifies to:
\[
-\frac{n(n-1)(n-2)}{6}x^3 - n^2(n-1)x^3 - \frac{n^2(n+1)}{2}x^3 - \frac{n(n+1)(n+2)}{6}x^3 = -n(5n^2-1)x^3
\]

#### Final Expansion:
\[
\left(\frac{1 - x}{1 + x}\right)^n = 1 - 2nx + n(2n-1)x^2 - n(5n^2-1)x^3 + \cdots
\]

### Conclusion:
The expansion of \(\left(\frac{1 - x}{1 + x}\right)^n\) up to the \(x^3\) term is:
\[
1 - 2nx + n(2n-1)x^2 - n(5n^2-1)x^3
\]

Would you like any further clarifications or details on specific steps?

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Here are 5 related questions:
1. How does the binomial theorem work for negative exponents?
2. What is the general form of a binomial expansion for \((1 + ax)^n\)?
3. Can you expand \(\left(\frac{1+x}{1-x}\right)^n\) up to the \(x^3\) term?
4. What happens when \(n\) is a fraction or irrational in binomial expansions?
5. How does this expansion change when \(x \to 1\)?

**Tip:** Always check the validity range of a binomial expansion—typically valid for \(|x| < 1\) in such cases.

Learn how to expand the expression ((1-x)/(1+x))^n using binomial series expansion up to the x^3 term. This solution covers key concepts such as algebra, binomial expansion, and series expansion, with a detailed step-by-step process. Suitable for undergraduate students studying advanced algebra.

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