The middle term in the expansion of `((3)/(x^(2))-(x^(3))/(6))^(9)` is -

Options

A. `(189)/(8)x^(2), (21)/(16)x^(7)`
B. `(189)/(8)x^(2), - (21)/(16)x^(7)`
C. `-(189)/(8)x^(2), -(21)/(16)x^(7)`
D. None of these

Answer

To find the middle term in the expansion of \(\left(\frac{3}{x^2} - \frac{x^3}{6}\right)^9\), we will follow these steps:

### Step 1: Identify the values of \(n\), \(x\), and \(y\)
In the expression \(\left(\frac{3}{x^2} - \frac{x^3}{6}\right)^9\):
- \(n = 9\)
- \(x = \frac{3}{x^2}\)
- \(y = -\frac{x^3}{6}\)

### Step 2: Determine the middle term
Since \(n\) is odd (9), the middle term is given by the formula:
\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]
The middle term will be the \(\left(\frac{n+1}{2} + 1\right)^{th}\) term, which is the \(5^{th}\) term (since \(n+1 = 10\) and \(\frac{10}{2} = 5\)).

### Step 3: Calculate \(T_5\)
Using the formula for the \(r^{th}\) term:
\[ T_5 = \binom{9}{4} \left(\frac{3}{x^2}\right)^{9-4} \left(-\frac{x^3}{6}\right)^4 \]

### Step 4: Substitute values into the formula
Substituting the values:
\[ T_5 = \binom{9}{4} \left(\frac{3}{x^2}\right)^{5} \left(-\frac{x^3}{6}\right)^{4} \]

### Step 5: Calculate \(\binom{9}{4}\)
\[ \binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \]

### Step 6: Calculate \(\left(\frac{3}{x^2}\right)^{5}\) and \(\left(-\frac{x^3}{6}\right)^{4}\)
\[ \left(\frac{3}{x^2}\right)^{5} = \frac{3^5}{x^{10}} = \frac{243}{x^{10}} \]
\[ \left(-\frac{x^3}{6}\right)^{4} = \frac{x^{12}}{6^4} = \frac{x^{12}}{1296} \]

### Step 7: Combine the terms
Now substituting back into \(T_5\):
\[ T_5 = 126 \cdot \frac{243}{x^{10}} \cdot \frac{x^{12}}{1296} \]
\[ = 126 \cdot \frac{243}{1296} \cdot x^{2} \]

### Step 8: Simplify the coefficients
Calculating the coefficient:
\[ \frac{243}{1296} = \frac{1}{6} \]
Thus,
\[ T_5 = 126 \cdot \frac{1}{6} \cdot x^2 = 21x^2 \]

### Conclusion
The middle term in the expansion of \(\left(\frac{3}{x^2} - \frac{x^3}{6}\right)^9\) is:
\[ 21x^2 \]