`(x+(1)/(x))^(6)`

Explanation

To expand the expression \((x + \frac{1}{x})^6\) using the Binomial Theorem, we follow these steps:

### Step 1: Identify A, B, and n
In the expression \((A + B)^n\), we identify:
- \(A = x\)
- \(B = \frac{1}{x}\)
- \(n = 6\)

### Step 2: Write the Binomial Expansion Formula
The Binomial Theorem states that:
\[ (A + B)^n = \sum_{k=0}^{n} \binom{n}{k} A^{n-k} B^k \]
where \(\binom{n}{k}\) is the binomial coefficient.

### Step 3: Apply the Formula
Using the identified values, we can expand:
\[ (x + \frac{1}{x})^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} \left(\frac{1}{x}\right)^k \]

### Step 4: Simplify Each Term
Now we compute each term in the expansion:
- For \(k = 0\): \(\binom{6}{0} x^{6-0} \left(\frac{1}{x}\right)^0 = 1 \cdot x^6 = x^6\)
- For \(k = 1\): \(\binom{6}{1} x^{6-1} \left(\frac{1}{x}\right)^1 = 6 \cdot x^5 \cdot \frac{1}{x} = 6x^5\)
- For \(k = 2\): \(\binom{6}{2} x^{6-2} \left(\frac{1}{x}\right)^2 = 15 \cdot x^4 \cdot \frac{1}{x^2} = 15x^4\)
- For \(k = 3\): \(\binom{6}{3} x^{6-3} \left(\frac{1}{x}\right)^3 = 20 \cdot x^3 \cdot \frac{1}{x^3} = 20\)
- For \(k = 4\): \(\binom{6}{4} x^{6-4} \left(\frac{1}{x}\right)^4 = 15 \cdot x^2 \cdot \frac{1}{x^4} = \frac{15}{x^2}\)
- For \(k = 5\): \(\binom{6}{5} x^{6-5} \left(\frac{1}{x}\right)^5 = 6 \cdot x^1 \cdot \frac{1}{x^5} = \frac{6}{x^4}\)
- For \(k = 6\): \(\binom{6}{6} x^{6-6} \left(\frac{1}{x}\right)^6 = 1 \cdot 1 \cdot \frac{1}{x^6} = \frac{1}{x^6}\)

### Step 5: Combine All Terms
Now we combine all the terms:
\[ (x + \frac{1}{x})^6 = x^6 + 6x^5 + 15x^4 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6} \]

### Final Result
Thus, the expansion of \((x + \frac{1}{x})^6\) is:
\[ x^6 + 6x^5 + 15x^4 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6} \]