If `C_(0),C_(1), C_(2),...,C_(N)` denote the binomial
coefficients in the expansion of `(1 + x)^(n)` , then
`1^(3). C_(1)-2^(3). C_(3) - 4^(3) . C_(4) + ...+ (-1)^(n-1)n^(3) C_(n)=`

Options

A. 0
B. `n^(3) 2^(n)`
C. `n(n-1)(n-2)2^(n-3)`
D. `2^(n)`

Explanation

To solve the problem, we need to evaluate the expression:

\[ S = 1^3 \cdot C_1 - 2^3 \cdot C_3 - 4^3 \cdot C_4 + \ldots + (-1)^{n-1} n^3 \cdot C_n \]

where \( C_r \) are the binomial coefficients from the expansion of \( (1 + x)^n \).

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficients**:
The binomial coefficients \( C_r \) are given by:
\[ C_r = \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
for \( r = 0, 1, 2, \ldots, n \).

2. **Rewriting the Expression**:
We can rewrite the expression \( S \) as:
\[ S = \sum_{r=1}^{n} (-1)^{r-1} r^3 C_r \]

3. **Using the Binomial Theorem**:
The binomial theorem states that:
\[ (1 + x)^n = \sum_{r=0}^{n} C_r x^r \]
We can differentiate this expression to find relationships involving \( r \) and \( C_r \).

4. **Differentiating the Binomial Expansion**:
Differentiate \( (1 + x)^n \) three times:
\[ \frac{d}{dx}(1 + x)^n = n(1 + x)^{n-1} \]
\[ \frac{d^2}{dx^2}(1 + x)^n = n(n-1)(1 + x)^{n-2} \]
\[ \frac{d^3}{dx^3}(1 + x)^n = n(n-1)(n-2)(1 + x)^{n-3} \]

5. **Evaluating at \( x = -1 \)**:
We substitute \( x = -1 \) into the third derivative:
\[ \frac{d^3}{dx^3}(1 + x)^n \bigg|_{x=-1} = n(n-1)(n-2)(1 - 1)^{n-3} = 0 \]
This indicates that the sum of the coefficients multiplied by their respective powers results in zero when evaluated at \( x = -1 \).

6. **Conclusion**:
Since the expression sums to zero, we conclude that:
\[ S = 0 \]

### Final Answer:
\[ S = 0 \]