Find the coefficent of `x^(5)` in the expansion of `(1+x)^(3)(1-x)^(6)`.

Explanation

To find the coefficient of \( x^5 \) in the expansion of \( (1+x)^3(1-x)^6 \), we will follow these steps:

### Step 1: Expand the individual binomials

First, we will expand \( (1+x)^3 \) and \( (1-x)^6 \) using the Binomial Theorem.

The Binomial Theorem states that:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

For \( (1+x)^3 \):
\[ (1+x)^3 = \sum_{k=0}^{3} \binom{3}{k} x^k = \binom{3}{0} + \binom{3}{1} x + \binom{3}{2} x^2 + \binom{3}{3} x^3 \]
Calculating the coefficients:
- \( \binom{3}{0} = 1 \)
- \( \binom{3}{1} = 3 \)
- \( \binom{3}{2} = 3 \)
- \( \binom{3}{3} = 1 \)

Thus,
\[ (1+x)^3 = 1 + 3x + 3x^2 + x^3 \]

For \( (1-x)^6 \):
\[ (1-x)^6 = \sum_{k=0}^{6} \binom{6}{k} (-x)^k = \binom{6}{0} - \binom{6}{1} x + \binom{6}{2} x^2 - \binom{6}{3} x^3 + \binom{6}{4} x^4 - \binom{6}{5} x^5 + \binom{6}{6} x^6 \]
Calculating the coefficients:
- \( \binom{6}{0} = 1 \)
- \( \binom{6}{1} = 6 \)
- \( \binom{6}{2} = 15 \)
- \( \binom{6}{3} = 20 \)
- \( \binom{6}{4} = 15 \)
- \( \binom{6}{5} = 6 \)
- \( \binom{6}{6} = 1 \)

Thus,
\[ (1-x)^6 = 1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6 \]

### Step 2: Multiply the two expansions

Now we need to multiply \( (1+x)^3 \) and \( (1-x)^6 \):
\[ (1 + 3x + 3x^2 + x^3)(1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6) \]

We will find the terms that contribute to \( x^5 \).

### Step 3: Identify combinations that yield \( x^5 \)

1. \( 1 \) from \( (1+x)^3 \) and \( -6x^5 \) from \( (1-x)^6 \):
- Contribution: \( 1 \cdot (-6) = -6 \)

2. \( 3x \) from \( (1+x)^3 \) and \( 15x^4 \) from \( (1-x)^6 \):
- Contribution: \( 3 \cdot 15 = 45 \)

3. \( 3x^2 \) from \( (1+x)^3 \) and \( -20x^3 \) from \( (1-x)^6 \):
- Contribution: \( 3 \cdot (-20) = -60 \)

4. \( x^3 \) from \( (1+x)^3 \) and \( 15x^2 \) from \( (1-x)^6 \):
- Contribution: \( 1 \cdot 15 = 15 \)

### Step 4: Sum the contributions

Now, we sum the contributions:
\[ -6 + 45 - 60 + 15 = -6 \]

### Final Answer

Thus, the coefficient of \( x^5 \) in the expansion of \( (1+x)^3(1-x)^6 \) is \( \boxed{-6} \).

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