Question

Factorise, $8p^3-\frac{27}{p^3}$
(3 marks)

Answer 1

We know $a^3-b^3$ = $(a-b)(a^2+ab+b^2)$
(1 mark)

$8p^3-\frac{27}{p^3}$ $$=(2p{)}^3-{\left(\frac{3}{p}\right)}^3$$
Here, $a=2p$ and $b=\frac{3}{p}$
(1 mark)

$$\begin{array}{cc}\therefore 8p^3-\frac{27}{p^3}=& (2p-\frac{3}{p})\left[\right(2p{)}^2+\left(2p\right)\left(\frac{3}{p}\right)+{\left(\frac{3}{p}\right)}^2]\\ & \dots [\because a^3-b^3=(a-b\left)(a^2+ab+b^2)\right]\\ =& (2p-\frac{3}{p})(4p^2+6+\frac{9}{p^2})\end{array}$$
(1 mark)