The coefficient of `x^(6)` in the expansion of `(1-x)^(8)(1+x)^(12)` is equal to

Options

A. 168
B. `-8`
C. 28
D. 104

Answer

To find the coefficient of \( x^6 \) in the expansion of \( (1-x)^8(1+x)^{12} \), we can use the binomial theorem to expand both terms separately and then combine them.

### Step 1: Expand \( (1-x)^8 \)
Using the binomial theorem, we have:
\[ (1-x)^8 = \sum_{k=0}^{8} \binom{8}{k} (-1)^k x^k \]
The relevant terms for \( x^k \) where \( k \leq 6 \) are:
- \( \binom{8}{0} (-1)^0 x^0 = 1 \)
- \( \binom{8}{1} (-1)^1 x^1 = -8x \)
- \( \binom{8}{2} (-1)^2 x^2 = 28x^2 \)
- \( \binom{8}{3} (-1)^3 x^3 = -56x^3 \)
- \( \binom{8}{4} (-1)^4 x^4 = 70x^4 \)
- \( \binom{8}{5} (-1)^5 x^5 = -56x^5 \)
- \( \binom{8}{6} (-1)^6 x^6 = 28x^6 \)

### Step 2: Expand \( (1+x)^{12} \)
Using the binomial theorem again, we have:
\[ (1+x)^{12} = \sum_{j=0}^{12} \binom{12}{j} x^j \]
The relevant terms for \( x^j \) where \( j \leq 6 \) are:
- \( \binom{12}{0} x^0 = 1 \)
- \( \binom{12}{1} x^1 = 12x \)
- \( \binom{12}{2} x^2 = 66x^2 \)
- \( \binom{12}{3} x^3 = 220x^3 \)
- \( \binom{12}{4} x^4 = 495x^4 \)
- \( \binom{12}{5} x^5 = 792x^5 \)
- \( \binom{12}{6} x^6 = 924x^6 \)

### Step 3: Combine the expansions to find the coefficient of \( x^6 \)
To find the coefficient of \( x^6 \) in the product \( (1-x)^8(1+x)^{12} \), we need to consider the combinations of terms from both expansions that yield \( x^6 \):
1. From \( (1-x)^8 \), take \( x^6 \) and from \( (1+x)^{12} \), take \( x^0 \):
\[ \text{Coefficient} = \binom{8}{6} \cdot \binom{12}{0} = 28 \cdot 1 = 28 \]
2. From \( (1-x)^8 \), take \( x^5 \) and from \( (1+x)^{12} \), take \( x^1 \):
\[ \text{Coefficient} = \binom{8}{5} \cdot \binom{12}{1} = -56 \cdot 12 = -672 \]
3. From \( (1-x)^8 \), take \( x^4 \) and from \( (1+x)^{12} \), take \( x^2 \):
\[ \text{Coefficient} = \binom{8}{4} \cdot \binom{12}{2} = 70 \cdot 66 = 4620 \]
4. From \( (1-x)^8 \), take \( x^3 \) and from \( (1+x)^{12} \), take \( x^3 \):
\[ \text{Coefficient} = \binom{8}{3} \cdot \binom{12}{3} = -56 \cdot 220 = -12320 \]
5. From \( (1-x)^8 \), take \( x^2 \) and from \( (1+x)^{12} \), take \( x^4 \):
\[ \text{Coefficient} = \binom{8}{2} \cdot \binom{12}{4} = 28 \cdot 495 = 13860 \]
6. From \( (1-x)^8 \), take \( x^1 \) and from \( (1+x)^{12} \), take \( x^5 \):
\[ \text{Coefficient} = \binom{8}{1} \cdot \binom{12}{5} = -8 \cdot 792 = -6336 \]
7. From \( (1-x)^8 \), take \( x^0 \) and from \( (1+x)^{12} \), take \( x^6 \):
\[ \text{Coefficient} = \binom{8}{0} \cdot \binom{12}{6} = 1 \cdot 924 = 924 \]

### Step 4: Sum the coefficients
Now, we sum all the coefficients:
\[ 28 - 672 + 4620 - 12320 + 13860 - 6336 + 924 \]
Calculating this step-by-step:
1. \( 28 - 672 = -644 \)
2. \( -644 + 4620 = 3976 \)
3. \( 3976 - 12320 = -8344 \)
4. \( -8344 + 13860 = 5226 \)
5. \( 5226 - 6336 = -1110 \)
6. \( -1110 + 924 = -186 \)

Thus, the coefficient of \( x^6 \) in the expansion of \( (1-x)^8(1+x)^{12} \) is:
\[ \boxed{-186} \]