Factorise :
`a^(2) + (1)/(a^(2)) - 18`

Answer

To factorise the expression \( a^2 + \frac{1}{a^2} - 18 \), we can follow these steps:

### Step 1: Rewrite the expression
First, we rewrite the expression in a more manageable form:
\[ a^2 + \frac{1}{a^2} - 18 \]

### Step 2: Combine terms
We can consider \( a^2 + \frac{1}{a^2} \) as a single term. To facilitate the factorization, we can express \( -18 \) as \( -2 - 16 \):
\[ a^2 + \frac{1}{a^2} - 2 - 16 \]

### Step 3: Recognize a perfect square
Notice that \( a^2 + \frac{1}{a^2} - 2 \) can be rewritten using the identity \( x^2 + y^2 - 2xy = (x - y)^2 \):
\[ a^2 + \frac{1}{a^2} - 2 = \left(a - \frac{1}{a}\right)^2 \]
Thus, we can rewrite the expression as:
\[ \left(a - \frac{1}{a}\right)^2 - 16 \]

### Step 4: Apply the difference of squares
Now, we can recognize this as a difference of squares, which follows the identity \( x^2 - y^2 = (x - y)(x + y) \):
Let \( x = a - \frac{1}{a} \) and \( y = 4 \):
\[ \left(a - \frac{1}{a} - 4\right)\left(a - \frac{1}{a} + 4\right) \]

### Step 5: Final factorization
Thus, the factorized form of the original expression is:
\[ \left(a - \frac{1}{a} - 4\right)\left(a - \frac{1}{a} + 4\right) \]

### Summary of the factorization
The final answer is:
\[ \left(a - \frac{1}{a} - 4\right)\left(a - \frac{1}{a} + 4\right) \]

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