Question

Factorise the following:
(i) $1–64a^3–12a+48a^2$

(ii) $8p^3+\frac{12}{5}{p}^2+\frac{6}{25}p+\frac{1}{125}$

Answer 1

(i)
$1–64a^3–12a+48a^2=(1{)}^3–(4a{)}^3–3\times 1^2\times 4a+3\times 1\times (4a{)}^2$
$\left[Usingtheidentity\right(a–b{)}^3=a^3–b^3–3a^{2}b+3a{b}^2]$
$=(1–4a{)}^3=(1–4a\left)\right(1–4a\left)\right(1-4a)$

(ii)
$8p^3+\frac{12}{5}{p}^2+\frac{6}{25}p+\frac{1}{125}$
$=(2p{)}^3+3\times (2p{)}^2\times \frac{1}{5}+3\times \left(2p\right)\times {\left(\frac{1}{5}\right)}^2+{\left(\frac{1}{5}\right)}^3$
$[Usingtheidentity,(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2]$
= $(2p+\frac{1}{5}{)}^3$
$=(2p+\frac{1}{5})(2p+\frac{1}{5})(2p+\frac{1}{5})$