Factorise each of the following: <br/><br/>(i) $27 y^{3}+125 z^{3}$<br/><br/>(ii) $64 m^{3}-343 n^{3}$
Solution:
(i) $27 y^{3}+125 z^{3}$
$=(3 y)^{3}+(5 z)^{3}$
$=(3 y+5 z)\left[(3 y)^{2}+(5 z)^{2}-(3 y)(5 z)\right]$
$\left[\because a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right]$
$=(3 y+5 z)\left[9 y^{2}+25 z^{2}-15 y z\right]$
(ii) $64 m^{3}-343 n^{3}$
$=(4 m)^{3}-(7 n)^{3}$
$=(4 m-7 n)\left[(4 m)^{2}+(7 n)^{2}+(4 m)(7 n)\right]$
$\left[\because a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right]$
$=(4 m-7 n)\left[16 m^{2}+49 n^{2}+28 m n\right]$
(i) $27 y^{3}+125 z^{3}$
$=(3 y)^{3}+(5 z)^{3}$
$=(3 y+5 z)\left[(3 y)^{2}+(5 z)^{2}-(3 y)(5 z)\right]$
$\left[\because a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\right]$
$=(3 y+5 z)\left[9 y^{2}+25 z^{2}-15 y z\right]$
(ii) $64 m^{3}-343 n^{3}$
$=(4 m)^{3}-(7 n)^{3}$
$=(4 m-7 n)\left[(4 m)^{2}+(7 n)^{2}+(4 m)(7 n)\right]$
$\left[\because a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right]$
$=(4 m-7 n)\left[16 m^{2}+49 n^{2}+28 m n\right]$