Question

Factorise $\frac{64}{125}{a}^3-\frac{96}{25}{a}^2+\frac{48}{5}a-8$.

Answer 1

Step 1: Simplify the polynomial

$$\begin{gathered} \frac{64}{125}{a}^3-\frac{96}{25}{a}^2+\frac{48}{5}a-8 \\ =\frac{64a^3-480a^2+1200a-1000}{125} \\ =\frac{8}{125}\left(8a^3-60a^2+150a-125\right) \end{gathered}$$

Step 2: Factorise the polynomial

$$\begin{gathered} \frac{8}{125}\left(8a^3-60a^2+150a-125\right) \\ =\frac{8}{125}\left[\left(2a^3\right)-{\left(5\right)}^3-3\times {\left(2a\right)}^2\times 5+3\times 2a\times {\left(5\right)}^2\right] \end{gathered}$$

using the identity ${\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{)}}^{\mathbf{3}}\mathbf{=}{\mathit{x}}^{\mathbf{3}}\mathbf{-}{\mathit{y}}^{\mathbf{3}}\mathbf{-}\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{+}\mathbf{3}\mathit{x}{\mathit{y}}^{\mathbf{2}}$

$=\frac{8}{125}{(2a-5)}^3$

Hence the factors of $\frac{64}{125}{a}^3-\frac{96}{25}{a}^2+\frac{48}{5}a-8$ is $\frac{8}{125}{(2a-5)}^3$.