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

Answer

$\frac{64}{125}\text{a}^3-\frac{96}{25}\text{a}^2+\frac{48}{5}\text{a}-8$ $\Big(\frac{4}{5}\text{a}\Big)^3-(2)^3-3\Big(\frac{4}{5}\text{a}\Big)^2(2)+3\Big(\frac{4}{5}\text{a}\Big)(2)^2$ $=\Big(\frac{4}{5}\text{a}-2\Big)^3$ Hence, factorisatoin of $\frac{64}{125}\text{a}^3-\frac{96}{25}\text{a}^2+\frac{48}{5}\text{a}-8$ is $\Big(\frac{4}{5}\text{a}-2\Big)^3$

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