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

Options

A. 6
B. 5
C. 4
D. 3

Answer

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

### Step 1: Rewrite the expression
We start by rewriting \( (1 + x + x^2)^{-3} \) using the identity:
\[ 1 + x + x^2 = \frac{1 - x^3}{1 - x} \]
Thus, we can express our original function as:
\[ (1 + x + x^2)^{-3} = \left( \frac{1 - x^3}{1 - x} \right)^{-3} = (1 - x^3)^{-3} (1 - x)^{3} \]

### Step 2: Use the Binomial Theorem
Now we can expand both parts using the Binomial Theorem.

1. For \( (1 - x^3)^{-3} \):
Using the Binomial series expansion:
\[ (1 - x)^{-n} = \sum_{k=0}^{\infty} \binom{n+k-1}{k} x^k \]
we have:
\[ (1 - x^3)^{-3} = \sum_{k=0}^{\infty} \binom{3+k-1}{k} (x^3)^k = \sum_{k=0}^{\infty} \binom{2+k}{k} x^{3k} \]

2. For \( (1 - x)^3 \):
Using the Binomial expansion:
\[ (1 - x)^3 = \sum_{j=0}^{3} \binom{3}{j} (-1)^j x^j = 1 - 3x + 3x^2 - x^3 \]

### Step 3: Combine the expansions
Now we need to find the coefficient of \( x^6 \) in the product of these two expansions:
\[ (1 - 3x + 3x^2 - x^3) \cdot \left( \sum_{k=0}^{\infty} \binom{2+k}{k} x^{3k} \right) \]

We will look for combinations of terms that yield \( x^6 \):
- From \( 1 \) in \( (1 - x)^3 \), we need \( x^6 \) from \( (1 - x^3)^{-3} \): This corresponds to \( k=2 \) (since \( 3k = 6 \)).
- Coefficient: \( \binom{2+2}{2} = \binom{4}{2} = 6 \)

- From \( -3x \) in \( (1 - x)^3 \), we need \( x^5 \) from \( (1 - x^3)^{-3} \): This corresponds to \( k=1 \) (since \( 3k = 3 \)).
- Coefficient: \( \binom{2+1}{1} = \binom{3}{1} = 3 \)
- Contribution: \( -3 \times 3 = -9 \)

- From \( 3x^2 \) in \( (1 - x)^3 \), we need \( x^4 \) from \( (1 - x^3)^{-3} \): This corresponds to \( k=1 \) (since \( 3k = 3 \)).
- Coefficient: \( \binom{2+0}{0} = 1 \)
- Contribution: \( 3 \times 1 = 3 \)

- From \( -x^3 \) in \( (1 - x)^3 \), we need \( x^3 \) from \( (1 - x^3)^{-3} \): This corresponds to \( k=1 \) (since \( 3k = 3 \)).
- Coefficient: \( \binom{2+1}{1} = 3 \)
- Contribution: \( -1 \times 3 = -3 \)

### Step 4: Sum the contributions
Now we sum all the contributions:
\[ 6 - 9 + 3 - 3 = -3 \]

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