Question

Factorise.
$\text{(i)}125p^3+q^3$
$\text{(ii)}24a^3+81b^3$

Answer 1

$\text{i)}125p^3+q^3$
$=(5p{)}^3+q^3$
$\text{Here,}a=5p\text{and}b=q$
$\therefore 125p^3+q^3=(5p+q)\left[\right(5p{)}^2–\left(5p\right)\left(q\right)+q^2]$
$\dots [\because a^3+b^3=(a+b\left)\right(a^2–ab+b^2\left)\right]$
$=(5p+q)(25p^2–5pq+q^2)$
$\text{(ii)}24a^3+81b^3$ [Taking out the common factor 3]\)
$=3\left[\right(2a{)}^3+\left(3b{)}^3\right]$
$\text{Here,}A=2a\text{and}B=3b$
$\therefore 24a^3+81b^3$
$=3\left\{\right(2a+3b\left)\right[(2a{)}^2–(2a\left)\right(3b)+(3b{)}^2\left]\right\}$
$\dots [\because A^3+B^3=(A+B\left)\right(A^2–AB+B^2\left)\right]$
$=3(2a+3b)(4a^2–6ab+9b^2)$