Find the term independent of x in `(4x^3 + (7)/(x^2))^14`

Answer

To find the term independent of \( x \) in the expression \( (4x^3 + \frac{7}{x^2})^{14} \), we will use the Binomial Theorem.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our expression, \( a = 4x^3 \), \( b = \frac{7}{x^2} \), and \( n = 14 \). Therefore, the general term becomes:
\[ T_{r+1} = \binom{14}{r} (4x^3)^{14-r} \left(\frac{7}{x^2}\right)^r \]

2. **Simplify the General Term**:
We can simplify \( T_{r+1} \):
\[ T_{r+1} = \binom{14}{r} (4^{14-r} x^{3(14-r)}) \left(\frac{7^r}{x^{2r}}\right) \]
This simplifies to:
\[ T_{r+1} = \binom{14}{r} 4^{14-r} 7^r x^{3(14-r) - 2r} \]
\[ = \binom{14}{r} 4^{14-r} 7^r x^{42 - 5r} \]

3. **Find the Term Independent of \( x \)**:
To find the term independent of \( x \), we need to set the exponent of \( x \) to zero:
\[ 42 - 5r = 0 \]
Solving for \( r \):
\[ 5r = 42 \quad \Rightarrow \quad r = \frac{42}{5} = 8.4 \]

4. **Conclusion**:
Since \( r \) must be a non-negative integer and \( \frac{42}{5} \) is not an integer, there is no value of \( r \) that will give us a term independent of \( x \). Therefore, there is no term independent of \( x \) in the expansion.

### Final Answer:
The term independent of \( x \) does not exist in the expansion, hence the answer is 0.