Question:
Factorise
$\frac{64}{125} a^{3}-\frac{96}{25} a^{2}+\frac{48}{5} a-8$
Solution:
$\frac{64}{125} a^{3}-\frac{96}{25} a^{2}+\frac{48}{5} a-8=\left(\frac{4}{5} a\right)^{3}-(2)^{3}-3\left(\frac{4}{5} a\right)^{2}(2)+3\left(\frac{4}{5} a\right)(2)^{2}$
$=\left(\frac{4}{5} a-2\right)^{3}$
Hence, factorisation of $\frac{64}{125} a^{3}-\frac{96}{25} a^{2}+\frac{48}{5} a-8$ is $\left(\frac{4}{5} a-2\right)^{3}$.
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