Factorise: $27 x^{3}+y^{3}+z^{3}-9 x y z$

Solution:

It is known that,

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

$\therefore 27 x^{3}+y^{3}+z^{3}-9 x y z$

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

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

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