Find the seventh term from the beginning in the expansion of `(root3sqrt(2)+(1)/(root3sqrt(3)))^(n)`

Answer

To find the seventh term from the beginning in the expansion of \((\sqrt[3]{\sqrt{2}} + \frac{1}{\sqrt[3]{\sqrt{3}}})^n\), we will follow these steps:

### Step 1: Simplify the Terms
First, we simplify the terms in the expression.

- The first term \(A = \sqrt[3]{\sqrt{2}} = (\sqrt{2})^{1/3} = 2^{1/6}\).
- The second term \(B = \frac{1}{\sqrt[3]{\sqrt{3}}} = \frac{1}{(\sqrt{3})^{1/3}} = 3^{-1/6}\).

### Step 2: Identify the General Term
The general term in the binomial expansion of \((A + B)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} A^{n-r} B^r \]
where \(T_{r+1}\) is the \((r+1)\)th term.

### Step 3: Find the 7th Term
Since we want the 7th term, we set \(r = 6\) (because \(T_{7} = T_{r+1}\)):
\[ T_7 = \binom{n}{6} A^{n-6} B^6 \]

### Step 4: Substitute the Values of A and B
Substituting \(A\) and \(B\) into the formula:
\[ T_7 = \binom{n}{6} (2^{1/6})^{n-6} (3^{-1/6})^6 \]

### Step 5: Simplify the Expression
Now we simplify the expression:
\[ T_7 = \binom{n}{6} (2^{(n-6)/6}) (3^{-1}) \]
\[ T_7 = \binom{n}{6} \cdot \frac{2^{(n-6)/6}}{3} \]

### Final Answer
Thus, the seventh term from the beginning in the expansion is:
\[ T_7 = \frac{\binom{n}{6} \cdot 2^{(n-6)/6}}{3} \]