We are asked to find the coefficient of $x^6$ in the expansion of $(1 - x^2)(1 + x)^5$.

Step-by-step solution:

1. Expand $(1 + x)^5$ using the Binomial Theorem: The Binomial Theorem gives the expansion of $(1 + x)^5$ as: $(1 + x)^5 = \sum_{k=0}^{5} \binom{5}{k} x^k = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5$

2. Multiply by $(1 - x^2)$: Now, multiply each term of $(1 + x)^5$ by both $1$ and $-x^2$: $(1 - x^2)(1 + x)^5 = (1)(1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5) - (x^2)(1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5)$ Distribute and simplify each term: $= 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5 - (x^2 + 5x^3 + 10x^4 + 10x^5 + 5x^6 + x^7)$

3. Simplify the expression: Now, collect like terms: $= 1 + 5x + (10x^2 - x^2) + (10x^3 - 5x^3) + (5x^4 - 10x^4) + (x^5 - 10x^5) - 5x^6 - x^7$ $= 1 + 5x + 9x^2 + 5x^3 - 5x^4 - 9x^5 - 5x^6 - x^7$

4. Identify the coefficient of $x^6$: From the simplified expression, the term involving $x^6$ is $-5x^6$, so the coefficient is $-5$.

Final Answer:

The coefficient of $x^6$ is $\boxed{-5}$.

Would you like further clarification or details? Here are some related questions:

1. How is the Binomial Theorem applied to expand expressions?
2. Can you explain the process of combining like terms in polynomial expressions?
3. What is the general formula for finding coefficients in binomial expansions?
4. How would this problem change if the powers were different in the binomial expansion?
5. Can you explain how the subtraction of $x^2$ affects each term in the multiplication?

Tip: When expanding binomials with multiplication, break the process down by multiplying each term and grouping like powers to simplify the problem systematically.