What is the middle term in the expansion of `((xsqrt( y ))/( 3) - ( 3)/( y sqrt( x)))^12` ?

Options

A. A. ` C ( 12,7) x^(3) y^(-3)`
B. B. `C(12,6) x^(-3)y^(3)`
C. C. `C(12,7) x^(-3) y^(3)`
D. D. `C ( 12,6) x^(3) y^(-3)`

Answer

To find the middle term in the expansion of \(\left(\frac{x \sqrt{y}}{3} - \frac{3}{y \sqrt{x}}\right)^{12}\), we can follow these steps:

### Step 1: Identify the values of \(n\), \(a\), and \(b\)
In the expression \(\left(a + b\right)^{n}\), we have:
- \(n = 12\)
- \(a = \frac{x \sqrt{y}}{3}\)
- \(b = -\frac{3}{y \sqrt{x}}\)

### Step 2: Determine the middle term
Since \(n = 12\) is even, the middle term is given by the formula:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
where \(r = \frac{n}{2} = 6\). Therefore, the middle term is the \(7^{th}\) term, which corresponds to \(r = 6\).

### Step 3: Substitute values into the formula
Now we can substitute \(n\), \(r\), \(a\), and \(b\) into the formula:
\[ T_{7} = \binom{12}{6} \left(\frac{x \sqrt{y}}{3}\right)^{12-6} \left(-\frac{3}{y \sqrt{x}}\right)^{6} \]
This simplifies to:
\[ T_{7} = \binom{12}{6} \left(\frac{x \sqrt{y}}{3}\right)^{6} \left(-\frac{3}{y \sqrt{x}}\right)^{6} \]

### Step 4: Simplify the terms
Calculating each part:
1. \(\left(\frac{x \sqrt{y}}{3}\right)^{6} = \frac{x^6 y^3}{3^6}\)
2. \(\left(-\frac{3}{y \sqrt{x}}\right)^{6} = \frac{(-3)^6}{(y \sqrt{x})^{6}} = \frac{729}{y^6 x^3}\)

Now substituting these back into \(T_{7}\):
\[ T_{7} = \binom{12}{6} \cdot \frac{x^6 y^3}{729} \cdot \frac{729}{y^6 x^3} \]

### Step 5: Cancel out terms
The \(729\) cancels out:
\[ T_{7} = \binom{12}{6} \cdot \frac{x^6 y^3}{y^6 x^3} = \binom{12}{6} \cdot \frac{x^{6-3}}{y^{6-3}} = \binom{12}{6} \cdot \frac{x^3}{y^3} \]

### Step 6: Final expression
Thus, the middle term in the expansion is:
\[ T_{7} = \binom{12}{6} x^3 y^{-3} \]