Find the coefficient of `x^6` in `(1 +x+x^2)^(-3)`

Answer

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

### Step 1: Rewrite the Expression
We start with the expression:
\[ (1 + x + x^2)^{-3} \]
We can rewrite \( 1 + x + x^2 \) using the formula for the sum of a geometric series. However, in this case, we will use the binomial expansion directly.

### Step 2: Use the Binomial Theorem
We can express \( (1 + x + x^2)^{-3} \) using the generalized binomial theorem:
\[ (1 + u)^{-n} = \sum_{k=0}^{\infty} \binom{-n}{k} u^k \]
where \( u = x + x^2 \) and \( n = 3 \).

### Step 3: Substitute and Expand
Substituting \( u = x + x^2 \):
\[ (1 + (x + x^2))^{-3} = \sum_{k=0}^{\infty} \binom{-3}{k} (x + x^2)^k \]

### Step 4: Expand \( (x + x^2)^k \)
Next, we need to expand \( (x + x^2)^k \):
\[ (x + x^2)^k = \sum_{j=0}^{k} \binom{k}{j} x^j (x^2)^{k-j} = \sum_{j=0}^{k} \binom{k}{j} x^{j + 2(k-j)} = \sum_{j=0}^{k} \binom{k}{j} x^{2k - j} \]

### Step 5: Find the Coefficient of \( x^6 \)
We need to find the terms where \( 2k - j = 6 \). Rearranging gives us:
\[ j = 2k - 6 \]
This means \( j \) must be non-negative, so:
\[ 2k - 6 \geq 0 \implies k \geq 3 \]

### Step 6: Calculate Coefficients
Now we can find the coefficients for \( k = 3, 4, 5, \ldots \):
- For \( k = 3 \):
\[ j = 2(3) - 6 = 0 \implies \text{Coefficient} = \binom{3}{0} = 1 \]
- For \( k = 4 \):
\[ j = 2(4) - 6 = 2 \implies \text{Coefficient} = \binom{4}{2} = 6 \]
- For \( k = 5 \):
\[ j = 2(5) - 6 = 4 \implies \text{Coefficient} = \binom{5}{4} = 5 \]
- For \( k = 6 \):
\[ j = 2(6) - 6 = 6 \implies \text{Coefficient} = \binom{6}{6} = 1 \]

### Step 7: Combine Coefficients
Now, we need to combine these coefficients with the binomial coefficients from the expansion:
\[ \text{Coefficient of } x^6 = \binom{-3}{3} \cdot 1 + \binom{-3}{4} \cdot 6 + \binom{-3}{5} \cdot 5 + \binom{-3}{6} \cdot 1 \]
Calculating these:
- \( \binom{-3}{3} = -\frac{3 \cdot 2 \cdot 1}{3!} = -1 \)
- \( \binom{-3}{4} = \frac{-3 \cdot -4}{4!} = \frac{12}{24} = \frac{1}{2} \)
- \( \binom{-3}{5} = -\frac{3 \cdot 4 \cdot 5}{5!} = -\frac{60}{120} = -\frac{1}{2} \)
- \( \binom{-3}{6} = \frac{-3 \cdot -4 \cdot -5 \cdot -6}{6!} = -\frac{360}{720} = -\frac{1}{2} \)

### Final Calculation
Combining these:
\[ \text{Coefficient of } x^6 = -1 + 6 \cdot \frac{1}{2} - 5 \cdot \frac{1}{2} + 1 = -1 + 3 - 2 + 1 = 1 \]

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