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

Given:

\( 27 x^{3}+y^{3}+z^{3}-9 x y z \)

To do:

We have to factorize the given expression.

Solution:

We know that,

$a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)$

Therefore,

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

$= (3x + y + z)[(3x)^2 + y^2 + z^2 - (3x)(y) - (y)(z) - (z)(3x)]$

$= (3x + y + z)(9x^2 + y^2 + z^2 - 3xy - yz - 3xz)$

Hence, $27 x^{3}+y^{3}+z^{3}-9 x y z  =(3x + y + z)(9x^2 + y^2 + z^2 - 3xy - yz - 3xz)$.