If `a + (1)/(a)= 6 and a ne 0`, find
`a- (1)/(a)`

Answer

To solve the problem, we need to find the value of \( a - \frac{1}{a} \) given that \( a + \frac{1}{a} = 6 \).

### Step-by-step Solution:

1. **Start with the given equation:**
\[ a + \frac{1}{a} = 6 \]

2. **Square both sides:**
\[ \left( a + \frac{1}{a} \right)^2 = 6^2 \]
This simplifies to:
\[ a^2 + 2 \cdot a \cdot \frac{1}{a} + \frac{1}{a^2} = 36 \]
Since \( a \cdot \frac{1}{a} = 1 \), we have:
\[ a^2 + 2 + \frac{1}{a^2} = 36 \]

3. **Rearranging the equation:**
\[ a^2 + \frac{1}{a^2} = 36 - 2 \]
Thus:
\[ a^2 + \frac{1}{a^2} = 34 \]

4. **Now, we need to find \( a - \frac{1}{a} \). We will square this expression:**
\[ \left( a - \frac{1}{a} \right)^2 = a^2 - 2 \cdot a \cdot \frac{1}{a} + \frac{1}{a^2} \]
This simplifies to:
\[ \left( a - \frac{1}{a} \right)^2 = a^2 - 2 + \frac{1}{a^2} \]

5. **Substituting the value of \( a^2 + \frac{1}{a^2} \):**
\[ \left( a - \frac{1}{a} \right)^2 = (a^2 + \frac{1}{a^2}) - 2 \]
Substituting \( a^2 + \frac{1}{a^2} = 34 \):
\[ \left( a - \frac{1}{a} \right)^2 = 34 - 2 = 32 \]

6. **Taking the square root of both sides:**
\[ a - \frac{1}{a} = \sqrt{32} \]
Simplifying \( \sqrt{32} \):
\[ a - \frac{1}{a} = \sqrt{16 \cdot 2} = 4\sqrt{2} \]

### Final Answer:
\[ a - \frac{1}{a} = 4\sqrt{2} \]