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

Options

A. 10
B. 62
C. 6
D. 66

Answer

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

### Step 1: Square both sides of the equation
We start with the equation:
\[ x - \frac{1}{x} = 8 \]
Now, we square both sides:
\[ \left( x - \frac{1}{x} \right)^2 = 8^2 \]
This gives us:
\[ x^2 - 2 \cdot x \cdot \frac{1}{x} + \left( \frac{1}{x} \right)^2 = 64 \]

### Step 2: Simplify the equation
The term \( -2 \cdot x \cdot \frac{1}{x} \) simplifies to \(-2\):
\[ x^2 - 2 + \frac{1}{x^2} = 64 \]

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

### Step 4: Calculate the final value
Now, we calculate:
\[ x^2 + \frac{1}{x^2} = 66 \]

Thus, the value of \( x^2 + \frac{1}{x^2} \) is \( 66 \).

### Final Answer
\[ \text{The value of } x^2 + \frac{1}{x^2} \text{ is } 66. \]

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