Factorise :
`a^(4) - 7a^(2) + 1`

Answer

To factorise the expression \( a^4 - 7a^2 + 1 \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ a^4 - 7a^2 + 1 \]
We can rewrite it by adding and subtracting \( 2a^2 \):
\[ a^4 - 7a^2 + 2a^2 - 2a^2 + 1 \]
This allows us to regroup the terms.

### Step 2: Group the terms
Now, we can group the terms:
\[ (a^4 + 1) + (2a^2 - 9a^2) \]
This simplifies to:
\[ a^4 + 1 - 9a^2 \]

### Step 3: Recognize the structure
Next, we can recognize that \( a^4 \) can be expressed as \( (a^2)^2 \) and \( 9a^2 \) as \( (3a)^2 \):
\[ (a^2)^2 - 9a^2 + 1 \]

### Step 4: Complete the square
We can rewrite \( a^4 - 7a^2 + 1 \) as:
\[ (a^2 + 1)^2 - (3a)^2 \]
This is now in the form of \( A^2 - B^2 \), where \( A = a^2 + 1 \) and \( B = 3a \).

### Step 5: Apply the difference of squares formula
Using the difference of squares formula \( A^2 - B^2 = (A + B)(A - B) \):
\[ (a^2 + 1 + 3a)(a^2 + 1 - 3a) \]

### Step 6: Write the final factorised form
Thus, we have:
\[ (a^2 + 3a + 1)(a^2 - 3a + 1) \]

### Final Answer
The factorised form of \( a^4 - 7a^2 + 1 \) is:
\[ (a^2 + 3a + 1)(a^2 - 3a + 1) \]