If the seventh, terms from the beginning and the end in the expansion of `(root(3)(2)+1/(root(3)(3)))^(n)` are equal, then n is equal to
(i) 10
(ii) 11
(iii) 12
(iv) 13

Options

A. 10
B. 11
C. 12
D. 13

Answer

To solve the problem, we need to find the value of \( n \) such that the 7th term from the beginning and the end in the expansion of \( \left( \sqrt[3]{2} + \frac{1}{\sqrt[3]{3}} \right)^n \) are equal.

### Step-by-Step Solution:

1. **Identify the Terms**:
- The 7th term from the beginning in the expansion is given by the formula:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = \sqrt[3]{2} \), \( b = \frac{1}{\sqrt[3]{3}} \), and \( r = 6 \) (since we want the 7th term).
- Therefore, the 7th term from the beginning \( T_7 \) is:
\[ T_7 = \binom{n}{6} \left( \sqrt[3]{2} \right)^{n-6} \left( \frac{1}{\sqrt[3]{3}} \right)^6 \]

2. **Calculate \( T_7 \)**:
- Simplifying \( T_7 \):
\[ T_7 = \binom{n}{6} \left( \sqrt[3]{2} \right)^{n-6} \cdot \left( \frac{1}{\sqrt[3]{3}} \right)^6 = \binom{n}{6} \cdot \left( \sqrt[3]{2} \right)^{n-6} \cdot \left( \frac{1}{3^{2}} \right) \]
\[ T_7 = \binom{n}{6} \cdot \left( \sqrt[3]{2} \right)^{n-6} \cdot \frac{1}{9} \]

3. **7th Term from the End**:
- The 7th term from the end can be calculated using:
\[ L_{r+1} = \binom{n}{r} a^r b^{n-r} \]
- For the 7th term from the end \( L_7 \) (where \( r = 6 \)):
\[ L_7 = \binom{n}{6} \left( \frac{1}{\sqrt[3]{3}} \right)^{n-6} \left( \sqrt[3]{2} \right)^6 \]
\[ L_7 = \binom{n}{6} \cdot \left( \frac{1}{\sqrt[3]{3}} \right)^{n-6} \cdot \left( \sqrt[3]{2} \right)^6 \]

4. **Calculate \( L_7 \)**:
- Simplifying \( L_7 \):
\[ L_7 = \binom{n}{6} \cdot \left( \frac{1}{3^{(n-6)/3}} \right) \cdot \left( \sqrt[3]{2} \right)^6 \]
\[ L_7 = \binom{n}{6} \cdot \left( \sqrt[3]{2} \right)^6 \cdot \frac{1}{3^{(n-6)/3}} \]

5. **Set the Two Terms Equal**:
- Since \( T_7 = L_7 \):
\[ \binom{n}{6} \cdot \left( \sqrt[3]{2} \right)^{n-6} \cdot \frac{1}{9} = \binom{n}{6} \cdot \left( \sqrt[3]{2} \right)^6 \cdot \frac{1}{3^{(n-6)/3}} \]

6. **Cancel \( \binom{n}{6} \)**:
- Canceling \( \binom{n}{6} \) from both sides (assuming \( n \geq 6 \)):
\[ \left( \sqrt[3]{2} \right)^{n-6} \cdot \frac{1}{9} = \left( \sqrt[3]{2} \right)^6 \cdot \frac{1}{3^{(n-6)/3}} \]

7. **Rearranging the Equation**:
- Rearranging gives:
\[ \left( \sqrt[3]{2} \right)^{n-6} \cdot 3^{(n-6)/3} = 9 \cdot \left( \sqrt[3]{2} \right)^6 \]

8. **Equating Powers**:
- This leads to:
\[ 2^{(n-6)/3} = 9 \cdot 2^2 \]
- Simplifying gives:
\[ 2^{(n-6)/3} = 2^2 \cdot 3^2 \]
- Taking logarithms or equating exponents gives:
\[ \frac{n-6}{3} = 2 \quad \Rightarrow \quad n-6 = 6 \quad \Rightarrow \quad n = 12 \]

### Final Answer:
Thus, the value of \( n \) is \( 12 \).