The `4^(th)` term from the end in the expansion of `((x^(3) )/( 3) - (3)/( x^2))^(7)` is

Options

A. `315x`
B. `21x`
C. `105x`
D. None of these

Answer

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

### Step 1: Identify the parameters
In the expression \(\left(\frac{x^3}{3} - \frac{3}{x^2}\right)^7\), we have:
- \(n = 7\)
- \(x = \frac{x^3}{3}\)
- \(y = -\frac{3}{x^2}\)

### Step 2: Determine the term from the end
To find the 4th term from the end, we can use the relationship:
\[ \text{rth term from the end} = (n - r + 2) \text{th term from the beginning} \]
Here, since we want the 4th term from the end, we set \(r = 4\). Thus,
\[ \text{4th term from the end} = (7 - 4 + 2) \text{th term from the beginning} = 5 \text{th term from the beginning} \]

### Step 3: Use the binomial theorem
The general term (r+1) in the binomial expansion is given by:
\[ T_r = \binom{n}{r} x^{n-r} y^r \]
For the 5th term, we have \(r = 4\):
\[ T_5 = \binom{7}{4} \left(\frac{x^3}{3}\right)^{7-4} \left(-\frac{3}{x^2}\right)^4 \]

### Step 4: Calculate the coefficients and powers
Now we can calculate:
- \(\binom{7}{4} = \frac{7!}{4! \cdot 3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\)
- \(\left(\frac{x^3}{3}\right)^{3} = \frac{x^9}{27}\)
- \(\left(-\frac{3}{x^2}\right)^{4} = \frac{81}{x^8}\)

### Step 5: Combine the terms
Now substituting these values back into the term:
\[ T_5 = 35 \cdot \frac{x^9}{27} \cdot \frac{81}{x^8} \]
This simplifies to:
\[ T_5 = 35 \cdot \frac{81}{27} \cdot x^{9-8} = 35 \cdot 3 \cdot x = 105x \]

### Final Answer
Thus, the 4th term from the end in the expansion is:
\[ \boxed{105x} \]