If `a ne 0 and a - (1)/(a)= 3`, find :
`a^(2) + (1)/(a^(2))`

Answer

To solve the problem where \( a - \frac{1}{a} = 3 \) and we need to find \( a^2 + \frac{1}{a^2} \), we can follow these steps:

### Step 1: Square both sides of the equation
We start with the equation:
\[ a - \frac{1}{a} = 3 \]
Now, we square both sides:
\[ \left(a - \frac{1}{a}\right)^2 = 3^2 \]

### Step 2: Apply the square of a binomial formula
Using the formula \( (x - y)^2 = x^2 - 2xy + y^2 \), we have:
\[ \left(a - \frac{1}{a}\right)^2 = a^2 - 2a\left(\frac{1}{a}\right) + \left(\frac{1}{a}\right)^2 \]
This simplifies to:
\[ a^2 - 2 + \frac{1}{a^2} \]

### Step 3: Set the equation equal to 9
From Step 1, we know:
\[ a^2 - 2 + \frac{1}{a^2} = 9 \]

### Step 4: Rearrange the equation
Now, we can rearrange the equation to isolate \( a^2 + \frac{1}{a^2} \):
\[ a^2 + \frac{1}{a^2} - 2 = 9 \]
Adding 2 to both sides gives:
\[ a^2 + \frac{1}{a^2} = 9 + 2 \]

### Step 5: Simplify the result
Thus, we find:
\[ a^2 + \frac{1}{a^2} = 11 \]

### Final Answer
The value of \( a^2 + \frac{1}{a^2} \) is \( 11 \).

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