Factorise :
`125a^(3) + (1)/(8)`

Answer

To factorise the expression \( 125a^3 + \frac{1}{8} \), we can follow these steps:

### Step 1: Rewrite the expression
We can rewrite \( 125a^3 \) and \( \frac{1}{8} \) in terms of cubes:
\[ 125a^3 = (5a)^3 \quad \text{and} \quad \frac{1}{8} = \left(\frac{1}{2}\right)^3 \]
So, we can rewrite the expression as:
\[ (5a)^3 + \left(\frac{1}{2}\right)^3 \]

### Step 2: Apply the sum of cubes formula
The sum of cubes can be factored using the formula:
\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]
Here, let \( x = 5a \) and \( y = \frac{1}{2} \).

### Step 3: Substitute \( x \) and \( y \) into the formula
Using the sum of cubes formula, we substitute \( x \) and \( y \):
\[ (5a + \frac{1}{2})\left((5a)^2 - (5a)(\frac{1}{2}) + \left(\frac{1}{2}\right)^2\right) \]

### Step 4: Simplify the expression
Now, we simplify the second part:
1. Calculate \( (5a)^2 = 25a^2 \)
2. Calculate \( (5a)(\frac{1}{2}) = \frac{5a}{2} \)
3. Calculate \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)

Putting it all together:
\[ (5a + \frac{1}{2})\left(25a^2 - \frac{5a}{2} + \frac{1}{4}\right) \]

### Step 5: Write the final factorised form
Thus, the factorised form of \( 125a^3 + \frac{1}{8} \) is:
\[ (5a + \frac{1}{2})\left(25a^2 - \frac{5a}{2} + \frac{1}{4}\right) \]

### Summary of the factorised expression
The final answer is:
\[ (5a + \frac{1}{2})\left(25a^2 - \frac{5a}{2} + \frac{1}{4}\right) \]

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