Find the sum of the coefficients of `x^(32)` and `x^(-18)` in the expansion of
`(2x^(3)-(3)/(x^(2)))^(14)`.

Answer

To find the sum of the coefficients of \( x^{32} \) and \( x^{-18} \) in the expansion of \( (2x^3 - \frac{3}{x^2})^{14} \), we can use the Binomial Theorem.

### Step-by-step Solution:

1. **Identify the General Term**:
The general term in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = 2x^3 \), \( b = -\frac{3}{x^2} \), and \( n = 14 \). Thus, the general term becomes:
\[ T_{r+1} = \binom{14}{r} (2x^3)^{14-r} \left(-\frac{3}{x^2}\right)^r \]

2. **Simplify the General Term**:
Simplifying \( T_{r+1} \):
\[ T_{r+1} = \binom{14}{r} (2^{14-r} x^{3(14-r)}) \left(-3^r x^{-2r}\right) \]
This simplifies to:
\[ T_{r+1} = \binom{14}{r} (-1)^r 2^{14-r} 3^r x^{42 - 5r} \]
where \( 42 - 5r \) is the power of \( x \).

3. **Finding the Coefficient of \( x^{32} \)**:
Set the exponent of \( x \) equal to 32:
\[ 42 - 5r = 32 \]
Solving for \( r \):
\[ 5r = 42 - 32 \implies 5r = 10 \implies r = 2 \]
Now, substitute \( r = 2 \) into the general term to find the coefficient:
\[ T_{3} = \binom{14}{2} (-1)^2 2^{14-2} 3^2 \]
Calculate:
\[ \binom{14}{2} = \frac{14 \times 13}{2 \times 1} = 91 \]
So,
\[ T_{3} = 91 \cdot 2^{12} \cdot 9 = 819 \cdot 2^{12} \]

4. **Finding the Coefficient of \( x^{-18} \)**:
Set the exponent of \( x \) equal to -18:
\[ 42 - 5m = -18 \]
Solving for \( m \):
\[ 5m = 42 + 18 \implies 5m = 60 \implies m = 12 \]
Substitute \( m = 12 \) into the general term to find the coefficient:
\[ T_{13} = \binom{14}{12} (-1)^{12} 2^{14-12} 3^{12} \]
Calculate:
\[ \binom{14}{12} = \binom{14}{2} = 91 \]
So,
\[ T_{13} = 91 \cdot 2^2 \cdot 3^{12} = 91 \cdot 4 \cdot 3^{12} \]

5. **Sum of Coefficients**:
Now, we need to find the sum of the coefficients of \( x^{32} \) and \( x^{-18} \):
\[ \text{Sum} = 819 \cdot 2^{12} + 91 \cdot 4 \cdot 3^{12} \]

### Final Calculation:
- \( 819 \cdot 2^{12} \) and \( 91 \cdot 4 \cdot 3^{12} \) can be calculated separately to get the final answer.