Formula use: \left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3

Explanation

Here, given expression is : 2x – 3y – z
Now, Find the cubes of :
\{\left(2x-3y\right)-z\}^3
\rightarrow\left(2x-3y\right)^3-3.\left(2x-3y\right)^2.z+3.\left(2x-3y\right).\left(z\right)^2-\left(z\right)^3
\rightarrow\{\left(2x\right)^3-3.\left(2x\right)^2.3y+3.2x.\left(3y\right)^2-\left(3y\right)^3\}-3.\{\left(2x\right)^2-2.2x.3y+\left(3y\right)^2\}.z+3.\left(2x-3y\right).z^2-z^3
\rightarrow\left({8x}^3-36x^2y+54xy^2-{9y}^3\right)-3.\left(4x^2-12xy+9y^2\right).z+6{\rm xz}^2-9yz^2-z^3
\rightarrow {8x}^3-36x^2y+54xy^2-{9y}^3-12x^2z+36xyz-27y^2z+6{\rm xz}^2-9yz^2-z^3
Ans: {8x}^3-36x^2y+54xy^2-{9y}^3-12x^2z+36xyz-27y^2z+6{\rm xz}^2-9yz^2-z^3\ .