Factorise
$8 x^{3}+27 y^{3}+125 z^{3}-90 x y z$
Without actually calculating the cubes, find the value of each of the following
$(14)^{3}+(27)^{3}-(41)^{3}$
The value of $249^{2}-248^{2}$ is
On dividing $x^{3}+a x^{2}+19 x+20$ by $(x+3),$ if the remainder is $a,$ then find the value of $a$.
Factorise $: 8 x^{3}+y^{3}-27 z^{3}+18 x y z$