Factorise :
`(8a^(3))/(27) - (b^(3))/(8)`

Answer

To factorise the expression \(\frac{8a^3}{27} - \frac{b^3}{8}\), we can follow these steps:

### Step 1: Rewrite the terms
First, we rewrite the constants in the expression:
\[ \frac{8a^3}{27} = \frac{(2^3)(a^3)}{3^3} \quad \text{and} \quad \frac{b^3}{8} = \frac{b^3}{2^3} \]
So the expression becomes:
\[ \frac{(2a)^3}{3^3} - \frac{b^3}{2^3} \]

### Step 2: Identify the cubes
Now we can express the entire expression as:
\[ \left(\frac{2a}{3}\right)^3 - \left(\frac{b}{2}\right)^3 \]

### Step 3: Apply the difference of cubes formula
We can use the formula for the difference of cubes:
\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \]
Here, let \(x = \frac{2a}{3}\) and \(y = \frac{b}{2}\). Thus, we have:
\[ \left(\frac{2a}{3} - \frac{b}{2}\right)\left(\left(\frac{2a}{3}\right)^2 + \left(\frac{2a}{3}\right)\left(\frac{b}{2}\right) + \left(\frac{b}{2}\right)^2\right) \]

### Step 4: Calculate \(x^2\), \(xy\), and \(y^2\)
Now we calculate each part:
1. \(x^2 = \left(\frac{2a}{3}\right)^2 = \frac{4a^2}{9}\)
2. \(xy = \left(\frac{2a}{3}\right)\left(\frac{b}{2}\right) = \frac{2ab}{6} = \frac{ab}{3}\)
3. \(y^2 = \left(\frac{b}{2}\right)^2 = \frac{b^2}{4}\)

### Step 5: Combine the results
Putting it all together, we have:
\[ \frac{2a}{3} - \frac{b}{2} \quad \text{and} \quad \frac{4a^2}{9} + \frac{ab}{3} + \frac{b^2}{4} \]

### Step 6: Final expression
Thus, the factorised form of the expression is:
\[ \left(\frac{2a}{3} - \frac{b}{2}\right)\left(\frac{4a^2}{9} + \frac{ab}{3} + \frac{b^2}{4}\right) \]

### Summary of the solution:
The factorisation of \(\frac{8a^3}{27} - \frac{b^3}{8}\) is:
\[ \left(\frac{2a}{3} - \frac{b}{2}\right)\left(\frac{4a^2}{9} + \frac{ab}{3} + \frac{b^2}{4}\right) \]