(x^ 2 -1)^ n = x ^ (2n) - n * x ^ (2n - 2) +***

The expression you provided appears to be the beginning of a binomial expansion. Let's expand the expression \((x^2 - 1)^n\) using the binomial theorem.

### Binomial Theorem
The binomial theorem states that:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

In your case, \( a = x^2 \) and \( b = -1 \). Applying the binomial theorem:

\[
(x^2 - 1)^n = \sum_{k=0}^{n} \binom{n}{k} (x^2)^{n-k} (-1)^k
\]

This expands to:

\[
(x^2 - 1)^n = \sum_{k=0}^{n} \binom{n}{k} x^{2(n-k)} (-1)^k
\]

### Explicit Expansion
Let's expand the first few terms:

\[
(x^2 - 1)^n = \binom{n}{0} x^{2n} (-1)^0 + \binom{n}{1} x^{2(n-1)} (-1)^1 + \binom{n}{2} x^{2(n-2)} (-1)^2 + \cdots
\]

Simplifying:

\[
(x^2 - 1)^n = x^{2n} - \binom{n}{1} x^{2(n-1)} + \binom{n}{2} x^{2(n-2)} - \cdots
\]

This shows the general form of the expansion. The exact number of terms and their coefficients will depend on the value of \( n \).

Would you like a more detailed breakdown of the terms for a specific value of \( n \), or do you have any further questions?

### Related Questions:
1. How does the binomial theorem apply when expanding \((1 + x)^n\) instead of \((x^2 - 1)^n\)?
2. What happens when \( n \) is a non-integer in the binomial expansion?
3. Can you derive the general formula for the coefficient of \( x^{2(n-k)} \) in the expansion of \((x^2 - 1)^n\)?
4. How does the expansion of \((x^2 - 1)^n\) change if we replace \( x^2 \) with another function, like \( x^3 \)?
5. What are some applications of binomial expansions in calculus or other areas of mathematics?

### Tip:
When expanding binomials, it's helpful to write out the first few terms explicitly to identify patterns in coefficients and powers. This is especially useful when analyzing specific cases or simplifying complex expressions.

Learn how to expand the expression (x^2 - 1)^n using the binomial theorem with a detailed step-by-step solution. This problem covers the application of the binomial theorem in algebra, suitable for advanced high school students and beyond.

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