Factorise :
`a^(2)b^(2) + 8ab - 9`

Answer

To factorise the expression \( a^2b^2 + 8ab - 9 \), we will follow these steps:

### Step 1: Identify the expression
The given expression is:
\[ a^2b^2 + 8ab - 9 \]
This is a quadratic expression in terms of \( ab \).

### Step 2: Rewrite the middle term
We need to split the middle term \( 8ab \) into two terms such that their product equals the product of the first term \( a^2b^2 \) and the last term \(-9\).

We need two numbers that multiply to \( -9 \times 1 = -9 \) (since the coefficient of \( a^2b^2 \) is 1) and add up to \( 8 \). The numbers that satisfy this are \( 9 \) and \( -1 \).

So, we can rewrite \( 8ab \) as:
\[ 9ab - ab \]
Thus, the expression becomes:
\[ a^2b^2 + 9ab - ab - 9 \]

### Step 3: Group the terms
Now, we will group the terms:
\[ (a^2b^2 + 9ab) + (-ab - 9) \]

### Step 4: Factor out common terms from each group
From the first group \( (a^2b^2 + 9ab) \), we can factor out \( ab \):
\[ ab(ab + 9) \]
From the second group \( (-ab - 9) \), we can factor out \(-1\):
\[ -1(ab + 9) \]

Now, we have:
\[ ab(ab + 9) - 1(ab + 9) \]

### Step 5: Factor out the common binomial
Now, we can see that \( (ab + 9) \) is common in both terms:
\[ (ab + 9)(ab - 1) \]

### Final Answer
Thus, the factorised form of the expression \( a^2b^2 + 8ab - 9 \) is:
\[ (ab + 9)(ab - 1) \]

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