To find the coefficient of \( x^{12} \) in the expansion of \( (1 - x^2 + x^4)^3 (1 - x)^7 \), we will break down the problem into manageable steps.
### Step 1: Expand \( (1 - x^2 + x^4)^3 \)
Using the multinomial expansion, we can expand \( (1 - x^2 + x^4)^3 \). The general term in the expansion of \( (a + b + c)^n \) is given by:
\[
\frac{n!}{p!q!r!} a^p b^q c^r
\]
where \( p + q + r = n \). Here, \( a = 1 \), \( b = -x^2 \), \( c = x^4 \), and \( n = 3 \).
The general term will be:
\[
\frac{3!}{p!q!r!} (1)^p (-x^2)^q (x^4)^r
\]
This simplifies to:
\[
\frac{6}{p!q!r!} (-1)^q x^{2q + 4r}
\]
### Step 2: Identify the relevant terms
We need to find the terms where \( 2q + 4r \) can contribute to \( x^{12} \). Thus, we need:
\[
2q + 4r = 12
\]
We also have the constraint:
\[
p + q + r = 3
\]
From \( p + q + r = 3 \), we can express \( p \) as:
\[
p = 3 - q - r
\]
### Step 3: Solve for combinations of \( q \) and \( r \)
Substituting \( p \) into \( 2q + 4r = 12 \):
1. If \( r = 0 \):
\[
2q = 12 \implies q = 6 \quad (\text{not possible since } p + q + r = 3)
\]
2. If \( r = 1 \):
\[
2q + 4 = 12 \implies 2q = 8 \implies q = 4 \quad (\text{not possible})
\]
3. If \( r = 2 \):
\[
2q + 8 = 12 \implies 2q = 4 \implies q = 2 \implies p = 3 - 2 - 2 = -1 \quad (\text{not possible})
\]
4. If \( r = 3 \):
\[
2q + 12 = 12 \implies 2q = 0 \implies q = 0 \implies p = 0
\]
This gives us the valid combination \( (p, q, r) = (0, 0, 3) \).
### Step 4: Calculate the coefficient for \( (1 - x^2 + x^4)^3 \)
The term for \( (p, q, r) = (0, 0, 3) \) is:
\[
\frac{3!}{0!0!3!} (1)^0 (-x^2)^0 (x^4)^3 = 1 \cdot x^{12}
\]
Thus, the coefficient of \( x^{12} \) in \( (1 - x^2 + x^4)^3 \) is \( 1 \).
### Step 5: Expand \( (1 - x)^7 \)
Using the binomial theorem, the expansion of \( (1 - x)^7 \) is:
\[
\sum_{k=0}^{7} \binom{7}{k} (-1)^k x^k
\]
### Step 6: Find the coefficient of \( x^{12} \)
To find the coefficient of \( x^{12} \) in the product \( (1)(1 - x)^7 \), we need to find the coefficient of \( x^{12} \) in \( (1)(1 - x)^7 \). Since the highest power of \( x \) in \( (1 - x)^7 \) is \( x^7 \), there is no \( x^{12} \) term in this expansion.
### Conclusion
Thus, the coefficient of \( x^{12} \) in the entire expression \( (1 - x^2 + x^4)^3 (1 - x)^7 \) is:
\[
\text{Coefficient} = 0
\]
