Factorise

We know

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

We have,

$\frac{x^{3}}{216}-8 y^{3}=\left(\frac{x}{6}\right)^{3}-(2 y)^{3}$

So, $a=\frac{x}{6}, b=2 y$

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