Factorise `(64)/(125)a^3-(96)/(25)a^2+(48)/(5)a-8`

Answer

To factorise the polynomial \(\frac{64}{125}a^3 - \frac{96}{25}a^2 + \frac{48}{5}a - 8\), we can follow these steps:

### Step 1: Rewrite the terms in a more manageable form
We can express the coefficients as cubes:
\[ \frac{64}{125} = \left(\frac{4}{5}\right)^3 \quad \text{and} \quad 8 = 2^3 \]
Thus, we can rewrite the polynomial as:
\[ \left(\frac{4}{5}\right)^3 a^3 - 2^3 - \frac{96}{25}a^2 + \frac{48}{5}a \]

### Step 2: Identify the cube terms
Now we can rewrite the first and last terms:
\[ \left(\frac{4a}{5}\right)^3 - 2^3 - \frac{96}{25}a^2 + \frac{48}{5}a \]

### Step 3: Factor out common terms
Next, we can factor out \(-\frac{24a}{5}\) from the remaining terms:
\[ \left(\frac{4a}{5}\right)^3 - 2^3 - \frac{24a}{5} \left(4a/5 - 2\right) \]

### Step 4: Recognize the difference of cubes
Now we have:
\[ \left(\frac{4a}{5}\right)^3 - 2^3 - \frac{24a}{5} \left(4a/5 - 2\right) \]
This can be recognized as a difference of cubes:
\[ A^3 - B^3 - 3AB(A - B) \]
where \(A = \frac{4a}{5}\) and \(B = 2\).

### Step 5: Apply the difference of cubes formula
Using the formula \(A^3 - B^3 = (A - B)(A^2 + AB + B^2)\), we can factor:
\[ \left(\frac{4a}{5} - 2\right)\left(\left(\frac{4a}{5}\right)^2 + \left(\frac{4a}{5}\cdot 2\right) + 2^2\right) \]

### Step 6: Simplify the expression
Now we can simplify the second factor:
\[ \left(\frac{4a}{5}\right)^2 = \frac{16a^2}{25}, \quad \left(\frac{4a}{5}\cdot 2\right) = \frac{8a}{5}, \quad \text{and} \quad 2^2 = 4 \]
Thus, the second factor becomes:
\[ \frac{16a^2}{25} + \frac{8a}{5} + 4 \]

### Step 7: Combine the factors
The complete factorization is:
\[ \left(\frac{4a}{5} - 2\right)\left(\frac{16a^2}{25} + \frac{8a}{5} + 4\right) \]

### Final Answer
The factorised form of the polynomial is:
\[ \left(\frac{4a}{5} - 2\right)\left(\frac{16a^2}{25} + \frac{8a}{5} + 4\right) \]