If `x+1/(x)=8`, then the value of `(x-1/(x))^(2)` is

Options

A. 64
B. 60
C. 16
D. 62

Answer

To solve the problem where \( x + \frac{1}{x} = 8 \) and we need to find the value of \( \left( x - \frac{1}{x} \right)^2 \), we can follow these steps:

### Step 1: Start with the given equation
We have:
\[ x + \frac{1}{x} = 8 \]

### Step 2: Square both sides
To find \( \left( x - \frac{1}{x} \right)^2 \), we first square the left-hand side:
\[ \left( x + \frac{1}{x} \right)^2 = 8^2 \]
This gives us:
\[ x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 64 \]
Since \( x \cdot \frac{1}{x} = 1 \), we can simplify this to:
\[ x^2 + 2 + \frac{1}{x^2} = 64 \]

### Step 3: Rearrange the equation
Now, we can rearrange the equation to isolate \( x^2 + \frac{1}{x^2} \):
\[ x^2 + \frac{1}{x^2} = 64 - 2 \]
This simplifies to:
\[ x^2 + \frac{1}{x^2} = 62 \]

### Step 4: Use the identity for \( \left( x - \frac{1}{x} \right)^2 \)
We know that:
\[ \left( x - \frac{1}{x} \right)^2 = \left( x + \frac{1}{x} \right)^2 - 4 \cdot x \cdot \frac{1}{x} \]
Substituting the values we have:
\[ \left( x - \frac{1}{x} \right)^2 = (x^2 + \frac{1}{x^2}) - 2 \]
Now substituting \( x^2 + \frac{1}{x^2} = 62 \):
\[ \left( x - \frac{1}{x} \right)^2 = 62 - 2 \]
This gives us:
\[ \left( x - \frac{1}{x} \right)^2 = 60 \]

### Final Answer
Thus, the value of \( \left( x - \frac{1}{x} \right)^2 \) is:
\[ \boxed{60} \]