Expand (i) `((2x)/3-(3)/(2x))^(6)` , (ii) `(2/x-x/2)^(5)`

Answer

To solve the given problems using the Binomial Theorem, we will expand the expressions step by step.

### Part (i): Expand \(\left(\frac{2x}{3} - \frac{3}{2x}\right)^{6}\)

1. **Identify the terms**: Let \( a = \frac{2x}{3} \) and \( b = -\frac{3}{2x} \). The expression can be rewritten as \( (a + b)^6 \).

2. **Use the Binomial Theorem**: The Binomial Theorem states:
\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r} \]
Here, \( n = 6 \).

3. **Calculate each term**:
- **First term**: \( T_0 = \binom{6}{0} a^6 b^0 = 1 \cdot \left(\frac{2x}{3}\right)^6 \cdot 1 = \frac{64x^6}{729} \)
- **Second term**: \( T_1 = \binom{6}{1} a^5 b^1 = 6 \cdot \left(\frac{2x}{3}\right)^5 \cdot \left(-\frac{3}{2x}\right) = -\frac{6 \cdot 32x^5 \cdot 3}{243} = -\frac{576x^4}{243} \)
- **Third term**: \( T_2 = \binom{6}{2} a^4 b^2 = 15 \cdot \left(\frac{2x}{3}\right)^4 \cdot \left(-\frac{3}{2x}\right)^2 = 15 \cdot \frac{16x^4}{81} \cdot \frac{9}{4} = \frac{240x^4}{81} \)
- **Fourth term**: \( T_3 = \binom{6}{3} a^3 b^3 = 20 \cdot \left(\frac{2x}{3}\right)^3 \cdot \left(-\frac{3}{2x}\right)^3 = -\frac{540x^3}{27} = -20x^3 \)
- **Fifth term**: \( T_4 = \binom{6}{4} a^2 b^4 = 15 \cdot \left(\frac{2x}{3}\right)^2 \cdot \left(-\frac{3}{2x}\right)^4 = \frac{15 \cdot 4x^2 \cdot 81}{16} = \frac{2430x^2}{16} \)
- **Sixth term**: \( T_5 = \binom{6}{5} a^1 b^5 = 6 \cdot \left(\frac{2x}{3}\right)^1 \cdot \left(-\frac{3}{2x}\right)^5 = -\frac{6 \cdot 2x \cdot 243}{243 \cdot 32} = -\frac{486x}{32} \)
- **Seventh term**: \( T_6 = \binom{6}{6} a^0 b^6 = \left(-\frac{3}{2x}\right)^6 = \frac{729}{64x^6} \)

4. **Combine all terms**:
\[ \left(\frac{2x}{3} - \frac{3}{2x}\right)^{6} = \frac{64x^6}{729} - \frac{576x^4}{243} + \frac{240x^4}{81} - 20x^3 + \frac{2430x^2}{16} - \frac{486x}{32} + \frac{729}{64x^6} \]

### Part (ii): Expand \(\left(\frac{2}{x} - \frac{x}{2}\right)^{5}\)

1. **Identify the terms**: Let \( a = \frac{2}{x} \) and \( b = -\frac{x}{2} \). The expression can be rewritten as \( (a + b)^5 \).

2. **Use the Binomial Theorem**: Here, \( n = 5 \).

3. **Calculate each term**:
- **First term**: \( T_0 = \binom{5}{0} a^5 b^0 = 1 \cdot \left(\frac{2}{x}\right)^5 = \frac{32}{x^5} \)
- **Second term**: \( T_1 = \binom{5}{1} a^4 b^1 = 5 \cdot \left(\frac{2}{x}\right)^4 \cdot \left(-\frac{x}{2}\right) = -\frac{80}{x^3} \)
- **Third term**: \( T_2 = \binom{5}{2} a^3 b^2 = 10 \cdot \left(\frac{2}{x}\right)^3 \cdot \left(-\frac{x}{2}\right)^2 = \frac{40}{x} \)
- **Fourth term**: \( T_3 = \binom{5}{3} a^2 b^3 = -10 \cdot \left(\frac{2}{x}\right)^2 \cdot \left(-\frac{x}{2}\right)^3 = -\frac{10x^2}{8} = -\frac{5x^2}{4} \)
- **Fifth term**: \( T_4 = \binom{5}{4} a^1 b^4 = 5 \cdot \left(\frac{2}{x}\right)^1 \cdot \left(-\frac{x}{2}\right)^4 = -\frac{5 \cdot 16}{2x} = -\frac{80}{2x} = -\frac{40}{x} \)
- **Sixth term**: \( T_5 = \binom{5}{5} a^0 b^5 = \left(-\frac{x}{2}\right)^5 = -\frac{x^5}{32} \)

4. **Combine all terms**:
\[ \left(\frac{2}{x} - \frac{x}{2}\right)^{5} = \frac{32}{x^5} - \frac{80}{x^3} + \frac{40}{x} - \frac{5x^2}{4} - \frac{40}{x} - \frac{x^5}{32} \]

### Final Answers:
1. \(\left(\frac{2x}{3} - \frac{3}{2x}\right)^{6} = \frac{64x^6}{729} - \frac{576x^4}{243} + \frac{240x^4}{81} - 20x^3 + \frac{2430x^2}{16} - \frac{486x}{32} + \frac{729}{64x^6}\)
2. \(\left(\frac{2}{x} - \frac{x}{2}\right)^{5} = \frac{32}{x^5} - \frac{80}{x^3} + \frac{40}{x} - \frac{5x^2}{4} - \frac{40}{x} - \frac{x^5}{32}\)