Factorise:

$27 x^{3}-y^{3}-z^{3}-9 x y z$

$=(3 x)^{3}-y^{3}-z^{3}-3 \times(3 x) \times(-y) \times(-z)$

We know,

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

$a=3 x, b=-y, c=-z$

$(3 x)^{3}-y^{3}-z^{3}-3 \times(3 x) \times(-y) \times(-z)=(3 x-y-z)\left(9 x^{2}+y^{2}+z^{2}+3 x y-y z+3 x z\right)$