If 5th term of the expansion `(root(3)(x) - 1/x)^n` is independent of x then n =

Options

A. 16
B. 12
C. 8
D. 4

Answer

To solve the problem, we need to find the value of \( n \) such that the 5th term of the expansion \( \left(\sqrt[3]{x} - \frac{1}{x}\right)^n \) is independent of \( x \).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = \sqrt[3]{x} \) and \( b = -\frac{1}{x} \).

2. **Substituting Values**:
In our case, we have:
\[ T_{r+1} = \binom{n}{r} \left(\sqrt[3]{x}\right)^{n-r} \left(-\frac{1}{x}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{n}{r} (-1)^r \left(\sqrt[3]{x}\right)^{n-r} x^{-r} \]
Further simplifying gives:
\[ T_{r+1} = \binom{n}{r} (-1)^r \cdot 3^{\frac{n-r}{3}} \cdot x^{\frac{n-r}{3} - r} \]

3. **Finding the 5th Term**:
The 5th term corresponds to \( r = 4 \):
\[ T_5 = \binom{n}{4} (-1)^4 \cdot 3^{\frac{n-4}{3}} \cdot x^{\frac{n-4}{3} - 4} \]
This simplifies to:
\[ T_5 = \binom{n}{4} \cdot 3^{\frac{n-4}{3}} \cdot x^{\frac{n-4}{3} - 4} \]

4. **Condition for Independence of \( x \)**:
For the term to be independent of \( x \), the exponent of \( x \) must be zero:
\[ \frac{n-4}{3} - 4 = 0 \]

5. **Solving the Equation**:
Rearranging the equation gives:
\[ \frac{n-4}{3} = 4 \]
Multiplying both sides by 3:
\[ n - 4 = 12 \]
Adding 4 to both sides:
\[ n = 16 \]

### Conclusion:
The value of \( n \) such that the 5th term of the expansion \( \left(\sqrt[3]{x} - \frac{1}{x}\right)^n \) is independent of \( x \) is \( n = 16 \).