Factorise :
`9a^(2) + (1)/(9a^(2)) - 2-12a + (4)/(3a)`

Answer

To factorise the expression \( 9a^2 + \frac{1}{9a^2} - 2 - 12a + \frac{4}{3a} \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ 9a^2 + \frac{1}{9a^2} - 2 - 12a + \frac{4}{3a} \]
We can rewrite \( 9a^2 \) as \( (3a)^2 \) and \( \frac{1}{9a^2} \) as \( \left(\frac{1}{3a}\right)^2 \):
\[ (3a)^2 + \left(\frac{1}{3a}\right)^2 - 2 - 12a + \frac{4}{3a} \]

### Step 2: Combine the terms
Next, we can combine the terms involving \( 3a \) and \( \frac{1}{3a} \):
\[ (3a)^2 - 12a + \left(\frac{1}{3a}\right)^2 + \frac{4}{3a} - 2 \]

### Step 3: Group the terms
Now, we can group the terms:
\[ (3a)^2 - 12a + \left(\frac{1}{3a}\right)^2 + \frac{4}{3a} - 2 \]
Notice that \( (3a)^2 - 12a \) can be factored as:
\[ (3a - 6)^2 \]
And we can rewrite \( \left(\frac{1}{3a}\right)^2 + \frac{4}{3a} - 2 \).

### Step 4: Factor the quadratic
We can now treat the expression as a quadratic in terms of \( 3a \) and \( \frac{1}{3a} \):
\[ (3a - 2)^2 - 4 \]
This can be factored using the difference of squares:
\[ (3a - 2 - 2)(3a - 2 + 2) = (3a - 4)(3a) \]

### Step 5: Final factorization
Thus, the final factorization is:
\[ (3a - 4)(3a + 1) \]