Factorise :
`(a^(2) - a)(4a^(2) -4a - 5) - 6`

Answer

To factorise the expression \((a^{2} - a)(4a^{2} - 4a - 5) - 6\), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (a^{2} - a)(4a^{2} - 4a - 5) - 6 \]

### Step 2: Factor out common terms
Notice that we can factor out \(a\) from \(a^{2} - a\):
\[ a(a - 1)(4a^{2} - 4a - 5) - 6 \]

### Step 3: Simplify the second part
Now, we need to simplify the expression \(4a^{2} - 4a - 5 - 6\):
\[ 4a^{2} - 4a - 11 \]

### Step 4: Substitute \(t\)
Let \(t = a^{2} - a\). Then we rewrite the expression:
\[ t(4t - 11) \]

### Step 5: Expand and rearrange
Now, we expand this:
\[ 4t^{2} - 11t \]

### Step 6: Factor the quadratic
To factor \(4t^{2} - 11t\), we look for two numbers that multiply to \(4 \times -11 = -44\) and add to \(-11\). The numbers are \(-4\) and \(+1\).

### Step 7: Rewrite the quadratic
We can rewrite the quadratic as:
\[ 4t^{2} - 4t + t - 11 \]

### Step 8: Group the terms
Now, we group the terms:
\[ (4t^{2} - 4t) + (t - 11) \]

### Step 9: Factor by grouping
Factoring out common terms in each group gives us:
\[ 4t(t - 1) + 1(t - 11) \]

### Step 10: Factor out the common binomial
Now, we can factor out the common binomial \((t - 11)\):
\[ (t - 11)(4t + 1) \]

### Step 11: Substitute back for \(t\)
We substitute back \(t = a^{2} - a\):
\[ (a^{2} - a - 11)(4(a^{2} - a) + 1) \]

### Step 12: Final factorization
Thus, the final factorization is:
\[ (a^{2} - a - 11)(4a^{2} - 4a + 1) \]