Find the 8th term in the expansion of `((2x)/(3)-(3)/(5x))^(12)`

Explanation

To find the 8th term in the expansion of \(\left(\frac{2x}{3} - \frac{3}{5x}\right)^{12}\), we can use the Binomial Theorem. According to the Binomial Theorem, the \(r\)th term in the expansion of \((a + b)^n\) is given by:

\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

In our case, \(n = 12\), \(a = \frac{2x}{3}\), and \(b = -\frac{3}{5x}\). We need to find the 8th term, which corresponds to \(r = 7\) (since the first term corresponds to \(r = 0\)).

### Step-by-step Solution:

1. **Identify the values**:
- \(n = 12\)
- \(a = \frac{2x}{3}\)
- \(b = -\frac{3}{5x}\)
- \(r = 7\)

2. **Use the Binomial Theorem to find the 8th term**:
\[ T_{8} = \binom{12}{7} \left(\frac{2x}{3}\right)^{12-7} \left(-\frac{3}{5x}\right)^7 \]

3. **Calculate the binomial coefficient**:
\[ \binom{12}{7} = \binom{12}{5} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792 \]

4. **Calculate \(a^{n-r}\)**:
\[ \left(\frac{2x}{3}\right)^{5} = \frac{(2x)^5}{3^5} = \frac{32x^5}{243} \]

5. **Calculate \(b^r\)**:
\[ \left(-\frac{3}{5x}\right)^{7} = (-1)^{7} \cdot \frac{3^7}{(5x)^7} = -\frac{2187}{78125x^7} \]

6. **Combine the terms**:
\[ T_{8} = 792 \cdot \frac{32x^5}{243} \cdot \left(-\frac{2187}{78125x^7}\right) \]

7. **Simplify the expression**:
\[ T_{8} = 792 \cdot \frac{32 \cdot (-2187)}{243 \cdot 78125} \cdot \frac{x^5}{x^7} \]
\[ = 792 \cdot \frac{32 \cdot (-2187)}{243 \cdot 78125} \cdot \frac{1}{x^2} \]

8. **Calculate the numerical part**:
\[ 32 \cdot (-2187) = -69984 \]
\[ T_{8} = 792 \cdot \frac{-69984}{243 \cdot 78125} \cdot \frac{1}{x^2} \]

9. **Final calculation**:
\[ = \frac{-792 \cdot 69984}{1899075} \cdot \frac{1}{x^2} \]

### Final Result:
The 8th term in the expansion is:
\[ T_{8} = \frac{-792 \cdot 69984}{1899075} \cdot \frac{1}{x^2} \]