Find the coefficient of `x^2 `in `(7x^3 - (2)/(x^2))^9`

Answer

To find the coefficient of \( x^2 \) in the expression \( (7x^3 - \frac{2}{x^2})^9 \), we will follow these steps:

### Step 1: Identify the general term
The general term \( T_r \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = 7x^3 \), \( b = -\frac{2}{x^2} \), and \( n = 9 \). Thus, the general term becomes:
\[ T_r = \binom{9}{r} (7x^3)^{9-r} \left(-\frac{2}{x^2}\right)^r \]

### Step 2: Simplify the general term
Now we simplify \( T_r \):
\[ T_r = \binom{9}{r} (7^{9-r} x^{3(9-r)}) \left(-2^r \cdot x^{-2r}\right) \]
Combining the powers of \( x \):
\[ T_r = \binom{9}{r} 7^{9-r} (-2)^r x^{27 - 3r - 2r} \]
This simplifies to:
\[ T_r = \binom{9}{r} 7^{9-r} (-2)^r x^{27 - 5r} \]

### Step 3: Set the exponent of \( x \) to 2
We need the exponent of \( x \) to equal 2:
\[ 27 - 5r = 2 \]
Solving for \( r \):
\[ 27 - 2 = 5r \implies 25 = 5r \implies r = 5 \]

### Step 4: Substitute \( r \) back into the general term
Now we substitute \( r = 5 \) back into the general term:
\[ T_5 = \binom{9}{5} 7^{9-5} (-2)^5 x^{27 - 5 \cdot 5} \]
Calculating the components:
\[ T_5 = \binom{9}{5} 7^4 (-2)^5 x^2 \]

### Step 5: Calculate the coefficient
Now we calculate the coefficient:
\[ \binom{9}{5} = \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \]
Calculating \( 7^4 \):
\[ 7^4 = 2401 \]
Calculating \( (-2)^5 \):
\[ (-2)^5 = -32 \]
Thus, the coefficient becomes:
\[ 126 \times 2401 \times (-32) \]
Calculating:
\[ 126 \times 2401 = 302526 \]
Now multiplying by \(-32\):
\[ 302526 \times (-32) = -9670592 \]

### Final Answer
The coefficient of \( x^2 \) in \( (7x^3 - \frac{2}{x^2})^9 \) is:
\[ \boxed{-9670592} \]