Question

Factorise each of the following:
(i) $8a^3+b^3+12a^b+6a{b}^2$
(ii) $8a^3-b^3-12a^{2}b+6a{b}^2$
(iii) $27-125a^3-135a+225a^2$
(iv) $64a^3-27b^3-144a^{2}b+108a{b}^2$
(v) $27p^3-\frac{1}{216}-\frac{9}{2}{p}^2+\frac{1}{4}p$

Answer 1

We know that

$(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2$ and $(a-b{)}^3=a^3+b^3-3a^{2}b-3a{b}^2$

(i) $8a^3+b^3+12a^{2}b+6a{b}^2=(2a{)}^3+(b{)}^3+3\left(2a{)}^2\right(b)+3(2a\left)\right(b{)}^2=(2a+b{)}^3$

(ii) $8a^3+b^3-12a^{2}b-6a{b}^2=(2a{)}^3+(b{)}^3-3\left(2a{)}^2\right(b)-3(2a\left)\right(b{)}^2=(2a-b{)}^3$

(iii) $27-125a^3-135a+225a^2=(3{)}^3+(-5a{)}^3-3\left(3{)}^2\right(-5a)-3(3\left)\right(-5a{)}^2=(3-5a{)}^3$

(iv) $64a^3-27b^3-144a^{2}b+108a{b}^2=(4a{)}^3+(-3b{)}^3+3\left(4a{)}^2\right(-3b)+3(4a\left)\right(-3b{)}^2=(4a-3b{)}^3$

(v) $27p^3+\frac{1}{216}-\frac{9}{2}{p}^2+\frac{1}{4}p$

$=(3p{)}^3{(-\frac{1}{6})}^3+3(3p){(-\frac{1}{6})}^2+3(3p){(-\frac{1}{6})}^2$

$={(3p-\frac{1}{6})}^3$