Question

Factorise. (i) $125p^3+q^3$ [2 marks]
(ii) $24a^3+81b^3$ [2 marks]

Answer 1

(i) $125p^3+q^3$
$=(5p{)}^3+q^3$
Here, a = 5p and b = q
$\therefore 125p^3+q^3=(5p+q)\left[\right(5p{)}^2-\left(5p\right)\left(q\right)+q^2]$
...$[\because a^3+b^3=(a+b\left)\right(a^2-ab+b^2\left)\right]$
(1 mark)
$=(5p+q)(25p^2-5pq+q^2)$
(1 mark)

(ii) $24a^3+81b^3$
[Taking out the common factor 3]
$=3\left[\right(2a{)}^3+\left(3b{)}^3\right]$
Here, A =2a and B = 3b
$\therefore 24a^3+81b^3$
$=3\left\{\begin{array}{c}(2a+3b)\left[\right((2a{)}^2-(2a\left)\right(3b)+(3b{)}^2]\end{array}\right\}$
...$[\because A^3+B^3=(A+B\left)\right(A^2-AB+B^2\left)\right]$
(1 mark)
$=3(2a+3b)(4a^2-6ab+9b^2)$
(1 mark)