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

We have

$27 x ^{3}+ y ^{3}+ z ^{3}-9 xyz =(3 x )^{3}+( y )^{3}+( z )^{3}-3(3 x )( y )( z )$

$\therefore $ Using the identity $x ^{3}+ y ^{3}+ z ^{3}-3 xyz =( x + y + z )\left( x ^{2}+ y ^{2}+ z ^{2}- xy – yz – zx \right)$,  we have  $(3 x )^{3}+( y )^{3}+( z )^{3}-3(3 x )( y )( z ) $

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

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