Write down the general term in the expansion of `(x^(2)-y^(3))^6`.
(ii) Determine `4^(th)` term from the end in the expansion of `((x^(3))/(2)-(2)/(x^(2)))^(9),x ne0`
(iii) Find the coefficient of `x^(-2)` in the expansion of `(x+(1)/(x^(3)))^(11),x ne 0`

Answer

Let's solve the three parts of the question step by step.

### Part (i): General Term in the Expansion of \((x^2 - y^3)^6\)

1. **Identify the Binomial Expansion Formula**: The general term \(T_{r+1}\) in the expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

2. **Substitute Values**: Here, \(a = x^2\), \(b = -y^3\), and \(n = 6\).
\[ T_{r+1} = \binom{6}{r} (x^2)^{6-r} (-y^3)^r \]

3. **Simplify the General Term**:
\[ T_{r+1} = \binom{6}{r} (x^{12 - 2r}) (-1)^r (y^{3r}) \]
\[ T_{r+1} = \binom{6}{r} (-1)^r x^{12 - 2r} y^{3r} \]

**General Term**:
\[ T_{r+1} = \binom{6}{r} (-1)^r x^{12 - 2r} y^{3r} \]

### Part (ii): Determine the 4th Term from the End in the Expansion of \(\left(\frac{x^3}{2} - \frac{2}{x^2}\right)^9\)

1. **Identify the Total Number of Terms**: The total number of terms in the expansion is \(n + 1 = 9 + 1 = 10\).

2. **Find the 4th Term from the End**: The 4th term from the end corresponds to the \(10 - 4 + 1 = 7\)th term from the beginning.

3. **Use the General Term Formula**:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \(n = 9\), \(a = \frac{x^3}{2}\), and \(b = -\frac{2}{x^2}\).

4. **Calculate the 7th Term**:
\[ T_7 = \binom{9}{6} \left(\frac{x^3}{2}\right)^{9-6} \left(-\frac{2}{x^2}\right)^6 \]
\[ T_7 = \binom{9}{6} \left(\frac{x^3}{2}\right)^3 \left(-\frac{2}{x^2}\right)^6 \]
\[ T_7 = \binom{9}{6} \frac{x^9}{8} \cdot \frac{-64}{x^{12}} \]
\[ T_7 = \binom{9}{6} \cdot \frac{-64}{8} \cdot x^{-3} \]

5. **Calculate the Binomial Coefficient**:
\[ \binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]
\[ T_7 = 84 \cdot \frac{-64}{8} \cdot x^{-3} = 84 \cdot -8 \cdot x^{-3} = -672 x^{-3} \]

**4th Term from the End**:
\[ T_7 = -672 x^{-3} \]

### Part (iii): Find the Coefficient of \(x^{-2}\) in the Expansion of \(\left(x + \frac{1}{x^3}\right)^{11}\)

1. **General Term in the Expansion**:
\[ T_{r+1} = \binom{11}{r} x^{11-r} \left(\frac{1}{x^3}\right)^r \]
\[ T_{r+1} = \binom{11}{r} x^{11 - r - 3r} = \binom{11}{r} x^{11 - 4r} \]

2. **Set the Exponent Equal to -2**:
\[ 11 - 4r = -2 \]
\[ 11 + 2 = 4r \implies 13 = 4r \implies r = \frac{13}{4} \]

3. **Check for Validity**: Since \(r\) must be an integer, \(r = \frac{13}{4}\) is not valid.

**Coefficient of \(x^{-2}\)**:
\[ \text{Coefficient} = 0 \]

### Summary of Solutions:
1. General Term: \(T_{r+1} = \binom{6}{r} (-1)^r x^{12 - 2r} y^{3r}\)
2. 4th Term from the End: \(T_7 = -672 x^{-3}\)
3. Coefficient of \(x^{-2}\): 0