Expand `(x+(1)/(x) )^(6). ( x ne 0)`

Answer

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 \((x + \frac{1}{x})^6\), we identify:
- \(a = x\)
- \(b = \frac{1}{x}\)
- \(n = 6\)

### Step 2: Apply the Binomial Theorem
According to the Binomial Theorem, the expansion of \((a + b)^n\) is given by:
\[ \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]
Thus, we can write:
\[ (x + \frac{1}{x})^6 = \sum_{r=0}^{6} \binom{6}{r} x^{6-r} \left(\frac{1}{x}\right)^r \]

### Step 3: Simplify Each Term
Now we will calculate each term of the expansion for \(r\) from 0 to 6:

1. For \(r = 0\):
\[ \binom{6}{0} x^{6-0} \left(\frac{1}{x}\right)^0 = 1 \cdot x^6 \cdot 1 = x^6 \]

2. For \(r = 1\):
\[ \binom{6}{1} x^{6-1} \left(\frac{1}{x}\right)^1 = 6 \cdot x^5 \cdot \frac{1}{x} = 6x^4 \]

3. For \(r = 2\):
\[ \binom{6}{2} x^{6-2} \left(\frac{1}{x}\right)^2 = 15 \cdot x^4 \cdot \frac{1}{x^2} = 15x^2 \]

4. For \(r = 3\):
\[ \binom{6}{3} x^{6-3} \left(\frac{1}{x}\right)^3 = 20 \cdot x^3 \cdot \frac{1}{x^3} = 20 \]

5. For \(r = 4\):
\[ \binom{6}{4} x^{6-4} \left(\frac{1}{x}\right)^4 = 15 \cdot x^2 \cdot \frac{1}{x^4} = \frac{15}{x^2} \]

6. For \(r = 5\):
\[ \binom{6}{5} x^{6-5} \left(\frac{1}{x}\right)^5 = 6 \cdot x^1 \cdot \frac{1}{x^5} = \frac{6}{x^4} \]

7. For \(r = 6\):
\[ \binom{6}{6} x^{6-6} \left(\frac{1}{x}\right)^6 = 1 \cdot 1 \cdot \frac{1}{x^6} = \frac{1}{x^6} \]

### Step 4: Combine All Terms
Now we combine all the terms from \(r = 0\) to \(r = 6\):
\[ (x + \frac{1}{x})^6 = x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6} \]

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