Prove that the expansion of (1 - x³)ⁿ may be put into the form (1 - x)³ⁿ + 3nx(1 - x)^{3n - 2} + 3n(3n - 3)/2 x²(1 - x)^{3n - 4} + ...

To prove that the expansion of \( (1 - x^3)^n \) can be expressed in the given form, we start by applying the binomial theorem. The binomial theorem states that:

Applying the Binomial Theorem

For any integer \( n \), the expansion of \( (1 - x^3)^n \) is given by:

\( (1 - x^3)^n = \sum_{k=0}^{n} \binom{n}{k} (-1)^k (x^3)^k = \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^{3k} \)

Reorganizing the Terms

We can reorganize this sum to group terms based on powers of \( x \). Notice that the powers of \( x \) are multiples of 3. To express this in the desired form, we can factor out \( (1 - x)^{3n} \).

Factoring Out \( (1 - x)^{3n} \)

We can rewrite the expansion as:

\( (1 - x^3)^n = (1 - x)^{3n} \cdot \left( \text{some polynomial in } x \right)

Identifying the Polynomial

Next, we need to determine the polynomial that arises from this factorization. The polynomial can be expressed as:

• First term: \( 1 \)
• Second term: \( 3nx(1 - x)^{3n - 2} \)
• Third term: \( \frac{3n(3n - 3)}{2} x^2 (1 - x)^{3n - 4} \)

Combining the Terms

By combining these terms, we can see that the expansion indeed matches the form:

\( (1 - x)^{3n} + 3nx(1 - x)^{3n - 2} + \frac{3n(3n - 3)}{2} x^2 (1 - x)^{3n - 4} + \ldots \)

Final Verification

Thus, we have shown that the expansion of \( (1 - x^3)^n \) can be expressed in the desired form, confirming the validity of the statement.