What is the coefficient of `x^(3) ` in `((3-2x))/((1+3x)^(3))` ?

Options

A. A. `-1728`
B. B. `-5040`
C. C. `-870`
D. D. `-918`

Answer

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

### Step 1: Rewrite the Expression
We start with the expression:
\[ \frac{3 - 2x}{(1 + 3x)^3} \]
This can be rewritten as:
\[ (3 - 2x) \cdot (1 + 3x)^{-3} \]

### Step 2: Use the Binomial Theorem
According to the Binomial Theorem, we can expand \( (1 + 3x)^{-3} \):
\[ (1 + 3x)^{-3} = \sum_{k=0}^{\infty} \binom{-3}{k} (3x)^k \]
The binomial coefficient \( \binom{-3}{k} \) can be calculated as:
\[ \binom{-3}{k} = (-1)^k \binom{3 + k - 1}{k} = (-1)^k \frac{(3 + k - 1)!}{(k!)(2)!} \]

### Step 3: Find the Terms of the Expansion
We need to find the terms of the expansion up to \( x^3 \):
- For \( k = 0 \):
\[ \binom{-3}{0} (3x)^0 = 1 \]
- For \( k = 1 \):
\[ \binom{-3}{1} (3x)^1 = -3(3x) = -9x \]
- For \( k = 2 \):
\[ \binom{-3}{2} (3x)^2 = \frac{(-3)(-4)}{2} (9x^2) = 6 \cdot 9x^2 = 54x^2 \]
- For \( k = 3 \):
\[ \binom{-3}{3} (3x)^3 = \frac{(-3)(-4)(-5)}{6} (27x^3) = -15 \cdot 27x^3 = -405x^3 \]

### Step 4: Combine with \( 3 - 2x \)
Now, we multiply each term by \( 3 - 2x \):
\[ (3 - 2x)(1 - 9x + 54x^2 - 405x^3) \]

### Step 5: Collect Terms for \( x^3 \)
To find the coefficient of \( x^3 \), we consider:
1. \( 3 \cdot (-405x^3) = -1215x^3 \)
2. \( -2x \cdot 54x^2 = -108x^3 \)

Adding these contributions together:
\[ -1215 - 108 = -1323 \]

### Final Result
Thus, the coefficient of \( x^3 \) in the expression \( \frac{3 - 2x}{(1 + 3x)^3} \) is:
\[ \boxed{-1323} \]