Let x be the `7^(th)` term from the beginning and y be the `7^(th)` term from the end in the expansion of `(3^(1//3)+(1)/(4^(1//3)))^(n`. If `y=12x` then the value of n is :

Options

A. 9
B. 8
C. 10
D. 11

Answer

To solve the problem, we need to find the value of \( n \) given the conditions about the 7th term from the beginning and the end in the expansion of \( (3^{1/3} + \frac{1}{4^{1/3}})^n \).

### Step-by-Step Solution:

1. **Identify the Terms**:
- The 7th term from the beginning, denoted as \( x \), can be expressed using the binomial theorem:
\[ x = T_7 = \binom{n}{6} (3^{1/3})^{n-6} \left(\frac{1}{4^{1/3}}\right)^6 \]
- The 7th term from the end, denoted as \( y \), can be expressed as:
\[ y = T_{n-6} = \binom{n}{n-6} (3^{1/3})^6 \left(\frac{1}{4^{1/3}}\right)^{n-6} \]

2. **Simplify the Terms**:
- Since \( \binom{n}{n-6} = \binom{n}{6} \), we can write:
\[ y = \binom{n}{6} (3^{1/3})^6 \left(\frac{1}{4^{1/3}}\right)^{n-6} \]
- Now, substituting the powers:
\[ x = \binom{n}{6} (3^{1/3})^{n-6} \cdot \left(\frac{1}{4^{1/3}}\right)^6 = \binom{n}{6} \cdot 3^{(n-6)/3} \cdot \frac{1}{4^2} = \binom{n}{6} \cdot 3^{(n-6)/3} \cdot \frac{1}{16} \]
\[ y = \binom{n}{6} \cdot 3^{2} \cdot \left(\frac{1}{4^{1/3}}\right)^{n-6} = \binom{n}{6} \cdot 9 \cdot \left(\frac{1}{4^{(n-6)/3}}\right) \]

3. **Set Up the Equation**:
- We are given that \( y = 12x \):
\[ \binom{n}{6} \cdot 9 \cdot \left(\frac{1}{4^{(n-6)/3}}\right) = 12 \cdot \left(\binom{n}{6} \cdot 3^{(n-6)/3} \cdot \frac{1}{16}\right) \]

4. **Cancel Common Terms**:
- Cancel \( \binom{n}{6} \) from both sides:
\[ 9 \cdot \frac{1}{4^{(n-6)/3}} = \frac{12 \cdot 3^{(n-6)/3}}{16} \]
- Simplifying the right side:
\[ 9 \cdot \frac{1}{4^{(n-6)/3}} = \frac{3 \cdot 3^{(n-6)/3}}{4} \]

5. **Cross-Multiply**:
- Cross-multiplying gives:
\[ 9 \cdot 4 = 3 \cdot 3^{(n-6)/3} \cdot 4^{(n-6)/3} \]
- This simplifies to:
\[ 36 = 3 \cdot (3 \cdot 4)^{(n-6)/3} = 3 \cdot 12^{(n-6)/3} \]

6. **Solve for \( n \)**:
- Dividing both sides by 3:
\[ 12 = 12^{(n-6)/3} \]
- Taking logarithm base 12:
\[ 1 = \frac{n-6}{3} \implies n - 6 = 3 \implies n = 9 \]

### Final Answer:
The value of \( n \) is \( 9 \).