If ab`ne` 0 and the sum of the coefficients of `x^(7)` and `x^(4)` in the expansion of `((x^(2))/a-b/x)^(11)` is zero, then ab=____

Answer

To solve the problem, we need to find the value of \( ab \) given that the sum of the coefficients of \( x^7 \) and \( x^4 \) in the expansion of \( \left( \frac{x^2}{a} - \frac{b}{x} \right)^{11} \) is zero.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term in the binomial expansion of \( (A + B)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} A^{n-r} B^r \]
In our case, \( A = \frac{x^2}{a} \) and \( B = -\frac{b}{x} \), and \( n = 11 \).

2. **Write the General Term**:
The general term for our expression becomes:
\[ T_{r+1} = \binom{11}{r} \left(\frac{x^2}{a}\right)^{11-r} \left(-\frac{b}{x}\right)^r \]
Simplifying this gives:
\[ T_{r+1} = \binom{11}{r} \frac{x^{2(11-r)}}{a^{11-r}} \cdot \frac{(-b)^r}{x^r} = \binom{11}{r} \frac{(-b)^r}{a^{11-r}} x^{22 - 3r} \]

3. **Find Coefficients for \( x^7 \) and \( x^4 \)**:
- For \( x^7 \):
\[ 22 - 3r = 7 \implies 3r = 15 \implies r = 5 \]
- For \( x^4 \):
\[ 22 - 3r = 4 \implies 3r = 18 \implies r = 6 \]

4. **Calculate Coefficients**:
- Coefficient of \( x^7 \) (when \( r = 5 \)):
\[ C_1 = \binom{11}{5} \frac{(-b)^5}{a^{6}} \]
- Coefficient of \( x^4 \) (when \( r = 6 \)):
\[ C_2 = \binom{11}{6} \frac{(-b)^6}{a^{5}} \]

5. **Set Up the Equation**:
The sum of the coefficients is given to be zero:
\[ C_1 + C_2 = 0 \]
Substituting the coefficients:
\[ \binom{11}{5} \frac{(-b)^5}{a^{6}} + \binom{11}{6} \frac{(-b)^6}{a^{5}} = 0 \]

6. **Simplify the Equation**:
Using the property \( \binom{11}{6} = \binom{11}{5} \):
\[ \binom{11}{5} \left( \frac{(-b)^5}{a^{6}} + \frac{(-b)^6}{a^{5}} \right) = 0 \]
Since \( \binom{11}{5} \neq 0 \), we can simplify:
\[ \frac{(-b)^5}{a^{6}} + \frac{(-b)^6}{a^{5}} = 0 \]

7. **Factor Out Common Terms**:
Factoring out \( \frac{(-b)^5}{a^{6}} \):
\[ \frac{(-b)^5}{a^{6}} \left( 1 - \frac{b}{a} \right) = 0 \]

8. **Solve for \( ab \)**:
Since \( ab \neq 0 \), we have:
\[ 1 - \frac{b}{a} = 0 \implies \frac{b}{a} = 1 \implies b = a \]
Therefore, substituting \( b = a \) into \( ab \):
\[ ab = a^2 \]
Since \( a \neq 0 \), we can set \( a = 1 \) (without loss of generality):
\[ ab = 1 \cdot 1 = 1 \]

### Final Answer:
\[ ab = 1 \]