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11. Factorise:     27x^3 + y^3 + z^3 - 9xyz

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Given is  27x^3 + y^3 + z^3 - 9xyz

Now, we know that 

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

Now, we can write  27x^3 + y^3 + z^3 - 9xyz  as 

\Rightarrow 27x^3 + y^3 + z^3 - 9xyz =(3x)^3+(y)^3+(z)^3-3.3x.y.z

Here, a= 3x , b = y \ \ and \ \ c = z

Therefore,

27x^3 + y^3 + z^3 - 9xyz =(3x+y+z)\left((3x)^2+(y)^2+(z)^2-3x.y-y.z-z.3x \right )

                                                 =(3x+y+z)\left(9x^2+y^2+z^2-3xy-yz-3zx \right )

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