Find `n` in the binomial `(2^(1/3)+1/(3^(1/3)))^n` , if the ration 7th term from the beginning to the 7 term from the end `1/6dot`

Explanation

To solve the problem, we need to find the value of \( n \) in the binomial expression \( (2^{1/3} + 3^{-1/3})^n \), given that the ratio of the 7th term from the beginning to the 7th term from the end is \( \frac{1}{6} \).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_r \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
For our case, \( a = 2^{1/3} \) and \( b = 3^{-1/3} \).

2. **Find the 7th Term from the Beginning**:
The 7th term from the beginning corresponds to \( r = 6 \):
\[ T_7 = \binom{n}{6} (2^{1/3})^{n-6} (3^{-1/3})^6 \]
Simplifying this, we get:
\[ T_7 = \binom{n}{6} \cdot 2^{(n-6)/3} \cdot 3^{-2} \]

3. **Find the 7th Term from the End**:
The 7th term from the end corresponds to \( r = n - 6 \):
\[ T_{n-5} = \binom{n}{n-6} (2^{1/3})^{6} (3^{-1/3})^{n-6} \]
Simplifying this, we get:
\[ T_{n-5} = \binom{n}{6} \cdot 2^{2} \cdot 3^{-(n-6)/3} \]

4. **Set Up the Ratio**:
According to the problem, the ratio of the 7th term from the beginning to the 7th term from the end is:
\[ \frac{T_7}{T_{n-5}} = \frac{1}{6} \]
Plugging in the expressions we derived:
\[ \frac{\binom{n}{6} \cdot 2^{(n-6)/3} \cdot 3^{-2}}{\binom{n}{6} \cdot 2^{2} \cdot 3^{-(n-6)/3}} = \frac{1}{6} \]

5. **Cancel Common Terms**:
The \( \binom{n}{6} \) cancels out:
\[ \frac{2^{(n-6)/3} \cdot 3^{-2}}{2^{2} \cdot 3^{-(n-6)/3}} = \frac{1}{6} \]

6. **Simplify the Equation**:
Rearranging gives:
\[ 2^{(n-6)/3 - 2} \cdot 3^{(n-6)/3 - 2} = \frac{1}{6} \]
This can be rewritten as:
\[ 6^{(n-12)/3} = \frac{1}{6} \]

7. **Equate the Exponents**:
Since \( \frac{1}{6} = 6^{-1} \), we have:
\[ \frac{n-12}{3} = -1 \]
Multiplying through by 3 gives:
\[ n - 12 = -3 \]
Therefore:
\[ n = 9 \]

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