Factorise : `a ^(2) + (1)/(a ^(2)) + 3 - 2a - (2)/(a)`

Options

A. `(a + (1)/(a) -a) (a - (1)/(a) + 1)`
B. `(a + (1)/(a) - 1) (a + (1)/(a) + 1 )`
C. `(a + (1)/(a) +1 ) (a + (1)/(a) + 1)`
D. `(a + (1)/(a) - 1) (a + (1)/(a) - 1)`

Answer

To factorize the expression \( a^2 + \frac{1}{a^2} + 3 - 2a - \frac{2}{a} \), we will follow these steps:

### Step 1: Rewrite the expression
Start with the original expression:
\[ a^2 + \frac{1}{a^2} + 3 - 2a - \frac{2}{a} \]

### Step 2: Group terms
Notice that \( a^2 + \frac{1}{a^2} \) can be rewritten using the identity \( (a + \frac{1}{a})^2 = a^2 + 2 + \frac{1}{a^2} \). We can rearrange the expression:
\[ = \left( a^2 + \frac{1}{a^2} \right) - 2a - \frac{2}{a} + 3 \]

### Step 3: Use the identity
We can express \( a^2 + \frac{1}{a^2} \) in terms of \( (a + \frac{1}{a})^2 \):
\[ = \left( (a + \frac{1}{a})^2 - 2 \right) - 2a - \frac{2}{a} + 3 \]
This simplifies to:
\[ = (a + \frac{1}{a})^2 - 2 - 2a - \frac{2}{a} + 3 \]
\[ = (a + \frac{1}{a})^2 - 2a - \frac{2}{a} + 1 \]

### Step 4: Combine constants
Now, combine the constants:
\[ = (a + \frac{1}{a})^2 - 2a - \frac{2}{a} + 1 \]
This can be rewritten as:
\[ = (a + \frac{1}{a})^2 - 2(a + \frac{1}{a}) + 1 \]

### Step 5: Factor the quadratic
Let \( x = a + \frac{1}{a} \). Then, we have:
\[ = x^2 - 2x + 1 \]
This factors to:
\[ = (x - 1)^2 \]

### Step 6: Substitute back
Substituting back \( x = a + \frac{1}{a} \):
\[ = (a + \frac{1}{a} - 1)^2 \]

### Final Result
Thus, the factorized form of the expression is:
\[ (a + \frac{1}{a} - 1)^2 \]