The coefficient of `x^(6)` in the expansion of `(1+x+x^(2)+x^(3))(1-x)^(6)` is

Options

A. `-10
B. 10
C. 9
D. `-9

Answer

To find the coefficient of \( x^6 \) in the expansion of \( (1 + x + x^2 + x^3)(1 - x)^6 \), we can follow these steps:

### Step 1: Expand \( (1 - x)^6 \)
Using the binomial theorem, we can expand \( (1 - x)^6 \):
\[ (1 - x)^6 = \sum_{k=0}^{6} \binom{6}{k} (-x)^k = \binom{6}{0} - \binom{6}{1} x + \binom{6}{2} x^2 - \binom{6}{3} x^3 + \binom{6}{4} x^4 - \binom{6}{5} x^5 + \binom{6}{6} x^6 \]
This gives us:
\[ (1 - x)^6 = 1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6 \]

### Step 2: Multiply by \( (1 + x + x^2 + x^3) \)
Now we need to multiply \( (1 + x + x^2 + x^3) \) with the expanded form of \( (1 - x)^6 \):
\[ (1 + x + x^2 + x^3)(1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6) \]

### Step 3: Find the coefficient of \( x^6 \)
We will find the contributions to \( x^6 \) from each term in \( (1 + x + x^2 + x^3) \):

1. **From \( 1 \)**:
- Coefficient of \( x^6 \) from \( (1 - x)^6 \) is \( \binom{6}{6} = 1 \).

2. **From \( x \)**:
- Coefficient of \( x^5 \) from \( (1 - x)^6 \) is \( -\binom{6}{5} = -6 \).

3. **From \( x^2 \)**:
- Coefficient of \( x^4 \) from \( (1 - x)^6 \) is \( \binom{6}{4} = 15 \).

4. **From \( x^3 \)**:
- Coefficient of \( x^3 \) from \( (1 - x)^6 \) is \( -\binom{6}{3} = -20 \).

Now, we can sum these contributions:
\[ \text{Coefficient of } x^6 = 1 - 6 + 15 - 20 \]

### Step 4: Calculate the final result
Calculating the above expression:
\[ 1 - 6 + 15 - 20 = 1 - 6 + 15 - 20 = -10 \]

Thus, the coefficient of \( x^6 \) in the expansion of \( (1 + x + x^2 + x^3)(1 - x)^6 \) is \(-10\).

### Final Answer
The coefficient of \( x^6 \) is \(-10\).