Factorise :
`(1)/(4)(a+b)^(2) - (9)/(16)(2a - b)^(2)`.

Answer

To factorise the expression \(\frac{1}{4}(a+b)^{2} - \frac{9}{16}(2a - b)^{2}\), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \frac{1}{4}(a+b)^{2} - \frac{9}{16}(2a - b)^{2} \]
We can rewrite \(\frac{1}{4}\) as \(\left(\frac{1}{2}\right)^{2}\) and \(\frac{9}{16}\) as \(\left(\frac{3}{4}\right)^{2}\). Thus, we have:
\[ \left(\frac{a+b}{2}\right)^{2} - \left(\frac{3(2a - b)}{4}\right)^{2} \]

### Step 2: Apply the difference of squares formula
We can now apply the difference of squares formula, which states that \(x^{2} - y^{2} = (x+y)(x-y)\). Here, let:
\[ x = \frac{a+b}{2} \quad \text{and} \quad y = \frac{3(2a - b)}{4} \]
So, we can write:
\[ \left(\frac{a+b}{2} + \frac{3(2a - b)}{4}\right)\left(\frac{a+b}{2} - \frac{3(2a - b)}{4}\right) \]

### Step 3: Simplify the first factor
Now we simplify the first factor:
\[ \frac{a+b}{2} + \frac{3(2a - b)}{4} \]
To add these fractions, we need a common denominator, which is 4:
\[ = \frac{2(a+b)}{4} + \frac{3(2a - b)}{4} = \frac{2a + 2b + 6a - 3b}{4} = \frac{8a - b}{4} \]

### Step 4: Simplify the second factor
Now we simplify the second factor:
\[ \frac{a+b}{2} - \frac{3(2a - b)}{4} \]
Again, using a common denominator of 4:
\[ = \frac{2(a+b)}{4} - \frac{3(2a - b)}{4} = \frac{2a + 2b - 6a + 3b}{4} = \frac{-4a + 5b}{4} \]

### Step 5: Combine the factors
Now we can combine the factors:
\[ \frac{(8a - b)(-4a + 5b)}{16} \]

### Final Answer
Thus, the factorised form of the expression is:
\[ \frac{(8a - b)(-4a + 5b)}{16} \]