Find the value

$=\frac{x^{3}}{6}-(2 y)^{3}$

$=(x / 6-2 y)\left((x / 6)^{2}+x / 6 \times 2 y+(2 y)^{2}\right)$

$\therefore\left[x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)\right]$

$=(x / 6-2 y)\left(x^{2} / 36+x y / 3+4 y^{2}\right)$

$\therefore x^{3} / 216-8 y^{3}=(x / 6-2 y)\left(x^{2} / 36+x y / 3+4 y^{2}\right)$