Question

Factorise : $\frac{64}{12}{a}^3-\frac{96}{25}{a}^2+\frac{48}{5}a-8$

Answer 1

Given : $\frac{64}{12}{a}^3-\frac{96}{25}{a}^2+\frac{48}{5}a-8$

According to the equation,

${(a+b)}^3=a^3+b^3+3b(a+b)$

Using the formula we can it write it as

$\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)\left(2\right)(\frac{4}{5}a-2)$

We can write it as

$={(\frac{4}{5}-2)}^3$

$\frac{64}{125}{a}^3-\frac{96}{25}{a}^2+\frac{48}{5}a-8$$=(\frac{4}{5}a-2)(\frac{4}{5}a-2)(\frac{4}{5}-2)$