The coefficient of `x^(7)` in the expansion of `(1-x-x^(2) + x^(3))^(6)` is :

Options

A. 144
B. -132
C. -144
D. 132

Answer

To find the coefficient of \( x^7 \) in the expansion of \( (1 - x - x^2 + x^3)^6 \), we can follow these steps:

### Step 1: Rewrite the expression
We can rewrite the expression as:
\[ (1 - x - x^2 + x^3)^6 = (1 - x)^6 (1 - x^2)^6 (1 + x)^6 \]

### Step 2: Use the binomial theorem
Using the binomial theorem, we know:
\[ (1 - x)^n = \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^k \]
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
\[ (1 - x^2)^n = \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^{2k} \]

### Step 3: Find the relevant coefficients
We need to find the coefficient of \( x^7 \) in the expansion. This can be done by considering the contributions from the different combinations of terms from each factor.

1. From \( (1 - x)^6 \), we can take \( x^k \) where \( k \) can be 1 to 6.
2. From \( (1 + x)^6 \), we can take \( x^m \) where \( m \) can be 0 to 6.
3. From \( (1 - x^2)^6 \), we can take \( x^{2j} \) where \( j \) can be 0 to 3.

We need to satisfy the equation:
\[ k + m + 2j = 7 \]

### Step 4: List the cases
We will consider the cases for \( j \) from 0 to 3:

- **Case 1**: \( j = 0 \) \(\Rightarrow k + m = 7\)
- Valid pairs: \( (6, 1), (5, 2), (4, 3), (3, 4), (2, 5), (1, 6) \)

- **Case 2**: \( j = 1 \) \(\Rightarrow k + m = 5\)
- Valid pairs: \( (5, 0), (4, 1), (3, 2), (2, 3), (1, 4), (0, 5) \)

- **Case 3**: \( j = 2 \) \(\Rightarrow k + m = 3\)
- Valid pairs: \( (3, 0), (2, 1), (1, 2), (0, 3) \)

- **Case 4**: \( j = 3 \) \(\Rightarrow k + m = 1\)
- Valid pairs: \( (1, 0), (0, 1) \)

### Step 5: Calculate the coefficients
Now we calculate the coefficients for each case:

1. **Case 1**:
- \( (6, 1) \): \( \binom{6}{6}(-1)^6 \cdot \binom{6}{1} = 1 \cdot 6 = 6 \)
- \( (5, 2) \): \( \binom{6}{5}(-1)^5 \cdot \binom{6}{2} = -6 \cdot 15 = -90 \)
- \( (4, 3) \): \( \binom{6}{4}(-1)^4 \cdot \binom{6}{3} = 15 \cdot 20 = 300 \)
- \( (3, 4) \): \( \binom{6}{3}(-1)^3 \cdot \binom{6}{4} = -20 \cdot 15 = -300 \)
- \( (2, 5) \): \( \binom{6}{2}(-1)^2 \cdot \binom{6}{5} = 15 \cdot 6 = 90 \)
- \( (1, 6) \): \( \binom{6}{1}(-1)^1 \cdot \binom{6}{6} = -6 \cdot 1 = -6 \)

Total for Case 1: \( 6 - 90 + 300 - 300 + 90 - 6 = 0 \)

2. **Case 2**:
- \( (5, 0) \): \( \binom{6}{5}(-1)^5 \cdot \binom{6}{0} = -6 \cdot 1 = -6 \)
- \( (4, 1) \): \( \binom{6}{4}(-1)^4 \cdot \binom{6}{1} = 15 \cdot 6 = 90 \)
- \( (3, 2) \): \( \binom{6}{3}(-1)^3 \cdot \binom{6}{2} = -20 \cdot 15 = -300 \)
- \( (2, 3) \): \( \binom{6}{2}(-1)^2 \cdot \binom{6}{3} = 15 \cdot 20 = 300 \)
- \( (1, 4) \): \( \binom{6}{1}(-1)^1 \cdot \binom{6}{4} = -6 \cdot 15 = -90 \)
- \( (0, 5) \): \( \binom{6}{0}(-1)^0 \cdot \binom{6}{5} = 1 \cdot 6 = 6 \)

Total for Case 2: \( -6 + 90 - 300 + 300 - 90 + 6 = 0 \)

3. **Case 3**:
- \( (3, 0) \): \( \binom{6}{3}(-1)^3 \cdot \binom{6}{0} = -20 \cdot 1 = -20 \)
- \( (2, 1) \): \( \binom{6}{2}(-1)^2 \cdot \binom{6}{1} = 15 \cdot 6 = 90 \)
- \( (1, 2) \): \( \binom{6}{1}(-1)^1 \cdot \binom{6}{2} = -6 \cdot 15 = -90 \)
- \( (0, 3) \): \( \binom{6}{0}(-1)^0 \cdot \binom{6}{3} = 1 \cdot 20 = 20 \)

Total for Case 3: \( -20 + 90 - 90 + 20 = 0 \)

4. **Case 4**:
- \( (1, 0) \): \( \binom{6}{1}(-1)^1 \cdot \binom{6}{0} = -6 \cdot 1 = -6 \)
- \( (0, 1) \): \( \binom{6}{0}(-1)^0 \cdot \binom{6}{1} = 1 \cdot 6 = 6 \)

Total for Case 4: \( -6 + 6 = 0 \)

### Final Calculation
Adding all contributions from all cases:
\[ 0 + 0 + 0 + 0 = 0 \]

### Conclusion
The coefficient of \( x^7 \) in the expansion of \( (1 - x - x^2 + x^3)^6 \) is \( \boxed{0} \).