(ii) Find the 4th term from the end in the expansion of `((3)/(x^2) - (x^3)/(6) )^7`.

Answer

To find the 4th term from the end in the expansion of \(\left(\frac{3}{x^2} - \frac{x^3}{6}\right)^7\), we will follow these steps:

### Step 1: Identify the number of terms in the expansion
The total number of terms in the expansion of \((a + b)^n\) is given by \(n + 1\). Here, \(n = 7\), so:
\[ \text{Total terms} = 7 + 1 = 8 \]

**Hint:** Remember that the number of terms in a binomial expansion is always \(n + 1\).

### Step 2: Determine the term we need
The 4th term from the end is equivalent to the 5th term from the beginning. In general, the \(k\)-th term in the expansion of \((a + b)^n\) is given by:
\[ T_k = \binom{n}{k-1} a^{n-(k-1)} b^{k-1} \]
Thus, for the 5th term:
\[ T_5 = \binom{7}{4} a^{7-4} b^{4} \]

**Hint:** The formula for the \(k\)-th term is crucial for finding specific terms in the expansion.

### Step 3: Substitute \(a\) and \(b\)
Here, \(a = \frac{3}{x^2}\) and \(b = -\frac{x^3}{6}\). Now we can substitute these values into the formula for \(T_5\):
\[ T_5 = \binom{7}{4} \left(\frac{3}{x^2}\right)^{3} \left(-\frac{x^3}{6}\right)^{4} \]

**Hint:** Make sure to keep track of the powers of \(a\) and \(b\) correctly.

### Step 4: Calculate the binomial coefficient
First, calculate \(\binom{7}{4}\):
\[ \binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \]

**Hint:** The binomial coefficient can be calculated using the factorial formula.

### Step 5: Calculate the powers of \(a\) and \(b\)
Now calculate \(a^3\) and \(b^4\):
\[ \left(\frac{3}{x^2}\right)^{3} = \frac{27}{x^6} \]
\[ \left(-\frac{x^3}{6}\right)^{4} = \frac{(-1)^4 x^{12}}{6^4} = \frac{x^{12}}{1296} \]

**Hint:** Pay attention to the signs and the powers when calculating \(a^n\) and \(b^n\).

### Step 6: Combine the results
Now, substitute these results back into the term:
\[ T_5 = 35 \cdot \frac{27}{x^6} \cdot \frac{x^{12}}{1296} \]

### Step 7: Simplify the expression
Now simplify:
\[ T_5 = 35 \cdot \frac{27}{1296} \cdot \frac{x^{12}}{x^6} = 35 \cdot \frac{27}{1296} \cdot x^{6} \]
Calculating \(35 \cdot \frac{27}{1296}\):
\[ T_5 = \frac{945}{1296} x^{6} \]

### Step 8: Final simplification
Now simplify \(\frac{945}{1296}\):
\[ \frac{945 \div 105}{1296 \div 105} = \frac{9}{12.8} = \frac{105}{144} \]
Thus, we have:
\[ T_5 = \frac{105}{144} x^{6} \]

### Final Answer:
The 4th term from the end in the expansion is:
\[ \frac{105}{144} x^{6} \]