`(root(6)3 sqrt2 + (1)/(root (3) 3))^(n) , (t _(7) "fron the " 1^(st))/( t _(7) "from the last")=1/6,` then the value of n is

Answer

To solve the problem, we need to find the value of \( n \) given the expression and the ratio of the 7th term from the beginning to the 7th term from the end. Let's break it down step by step.

### Step 1: Identify the General Term
For a binomial expression \( (x + y)^n \), the \( r \)-th term (or \( T_{r+1} \)) is given by:
\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]
In our case, we have \( x = \sqrt[6]{3} \) and \( y = \sqrt{2} + \frac{1}{\sqrt[3]{3}} \).

### Step 2: Find the 7th Term from the Beginning
The 7th term from the beginning is:
\[ T_7 = \binom{n}{6} x^{n-6} y^6 \]
Substituting \( x \) and \( y \):
\[ T_7 = \binom{n}{6} \left(\sqrt[6]{3}\right)^{n-6} \left(\sqrt{2} + \frac{1}{\sqrt[3]{3}}\right)^6 \]

### Step 3: Find the 7th Term from the End
The 7th term from the end is equivalent to the 7th term from the beginning of the expression \( (y + x)^n \). Therefore:
\[ T_7' = \binom{n}{6} y^{n-6} x^6 \]
Substituting \( y \) and \( x \):
\[ T_7' = \binom{n}{6} \left(\sqrt{2} + \frac{1}{\sqrt[3]{3}}\right)^{n-6} \left(\sqrt[6]{3}\right)^6 \]

### Step 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 given as:
\[ \frac{T_7}{T_7'} = \frac{1}{6} \]
Substituting the expressions for \( T_7 \) and \( T_7' \):
\[ \frac{\binom{n}{6} \left(\sqrt[6]{3}\right)^{n-6} \left(\sqrt{2} + \frac{1}{\sqrt[3]{3}}\right)^6}{\binom{n}{6} \left(\sqrt{2} + \frac{1}{\sqrt[3]{3}}\right)^{n-6} \left(\sqrt[6]{3}\right)^6} = \frac{1}{6} \]
The \( \binom{n}{6} \) cancels out.

### Step 5: Simplify the Equation
This simplifies to:
\[ \frac{\left(\sqrt[6]{3}\right)^{n-6}}{\left(\sqrt{2} + \frac{1}{\sqrt[3]{3}}\right)^{n-6}} \cdot \frac{\left(\sqrt{2} + \frac{1}{\sqrt[3]{3}}\right)^6}{\left(\sqrt[6]{3}\right)^6} = \frac{1}{6} \]
This leads to:
\[ \left(\sqrt[6]{3}\right)^{n-12} \cdot \left(\sqrt{2} + \frac{1}{\sqrt[3]{3}}\right)^{6-n+6} = \frac{1}{6} \]

### Step 6: Solve for \( n \)
Now, we can equate the powers of both sides:
\[ \frac{n-12}{6} = -1 \implies n - 12 = -6 \implies n = 6 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{10} \]