Write down and simplify
Write down and simplify 6th term in `((2x)/(3) + (3y)/(2))^9`

Answer

To find the sixth term in the expansion of \(\left(\frac{2x}{3} + \frac{3y}{2}\right)^9\), we will use the Binomial Theorem. The Binomial Theorem states that the \(r\)-th term in the expansion of \((a + b)^n\) is given by:

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

### Step-by-Step Solution:

1. **Identify the values**:
- Here, \(n = 9\), \(a = \frac{2x}{3}\), and \(b = \frac{3y}{2}\).
- We want to find the sixth term, which corresponds to \(r = 5\) (since the first term corresponds to \(r = 0\)).

2. **Write the formula for the sixth term**:
\[ T_{6} = \binom{9}{5} \left(\frac{2x}{3}\right)^{9-5} \left(\frac{3y}{2}\right)^5 \]

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

4. **Calculate the powers**:
- For \(\left(\frac{2x}{3}\right)^{4}\):
\[ \left(\frac{2x}{3}\right)^{4} = \frac{(2x)^{4}}{3^{4}} = \frac{16x^{4}}{81} \]
- For \(\left(\frac{3y}{2}\right)^{5}\):
\[ \left(\frac{3y}{2}\right)^{5} = \frac{(3y)^{5}}{2^{5}} = \frac{243y^{5}}{32} \]

5. **Combine everything**:
\[ T_{6} = 126 \cdot \frac{16x^{4}}{81} \cdot \frac{243y^{5}}{32} \]

6. **Simplify**:
- First, multiply the coefficients:
\[ 126 \cdot \frac{16 \cdot 243}{81 \cdot 32} \]
- Simplifying:
\[ \frac{16}{32} = \frac{1}{2}, \quad \frac{243}{81} = 3 \]
- Therefore:
\[ T_{6} = 126 \cdot \frac{1}{2} \cdot 3 = 126 \cdot \frac{3}{2} = 189 \]

7. **Final expression**:
\[ T_{6} = 189 x^{4} y^{5} \]

### Final Answer:
The sixth term in the expansion of \(\left(\frac{2x}{3} + \frac{3y}{2}\right)^9\) is:
\[ 189 x^{4} y^{5} \]