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. The Binomial Theorem states that:

\[ (p + q)^n = \sum_{r=0}^{n} \binom{n}{r} p^{n-r} q^r \]

### Step-by-Step Solution:

1. **Identify \(p\), \(q\), and \(n\)**:
- Here, \(p = \frac{2x}{3}\), \(q = -\frac{3}{5x}\), and \(n = 12\).

2. **Write the general term**:
- The general term \(T_{r+1}\) in the expansion is given by:
\[ T_{r+1} = \binom{n}{r} p^{n-r} q^r \]
- For our case, this becomes:
\[ T_{r+1} = \binom{12}{r} \left(\frac{2x}{3}\right)^{12-r} \left(-\frac{3}{5x}\right)^{r} \]

3. **Find the 8th term**:
- The 8th term corresponds to \(r = 7\) (since \(T_{r+1}\) means \(T_8\) corresponds to \(r = 7\)):
\[ T_8 = \binom{12}{7} \left(\frac{2x}{3}\right)^{12-7} \left(-\frac{3}{5x}\right)^{7} \]

4. **Calculate each component**:
- Calculate \(\binom{12}{7}\):
\[ \binom{12}{7} = \frac{12!}{7!(12-7)!} = \frac{12!}{7!5!} = 792 \]
- Calculate \(\left(\frac{2x}{3}\right)^{5}\):
\[ \left(\frac{2x}{3}\right)^{5} = \frac{(2^5)(x^5)}{3^5} = \frac{32x^5}{243} \]
- Calculate \(\left(-\frac{3}{5x}\right)^{7}\):
\[ \left(-\frac{3}{5x}\right)^{7} = \frac{(-3)^7}{(5x)^7} = \frac{-2187}{78125x^7} \]

5. **Combine the components**:
- Now substitute these values back into the expression for \(T_8\):
\[ T_8 = 792 \cdot \frac{32x^5}{243} \cdot \frac{-2187}{78125x^7} \]

6. **Simplify the expression**:
- Combine the constants:
\[ T_8 = 792 \cdot 32 \cdot \frac{-2187}{243 \cdot 78125} \cdot \frac{x^5}{x^7} \]
- Simplifying \(x^5/x^7 = \frac{1}{x^2}\):
\[ T_8 = -\frac{792 \cdot 32 \cdot 2187}{243 \cdot 78125} \cdot \frac{1}{x^2} \]

7. **Calculate the numerical value**:
- Calculate \(792 \cdot 32 = 25344\)
- Now calculate \(-\frac{25344 \cdot 2187}{243 \cdot 78125}\):
- This results in a final numerical value (after performing the calculations).

8. **Final answer**:
- After performing the calculations, we find:
\[ T_8 = -\frac{228096}{78125} \cdot \frac{1}{x^2} \]