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Question

Factorise 27x3+y3+z3-9xyz

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Solution

27x^3+y^3+z^3-9xyz

(3x)^3+y^3+z^3-9xyz

we have a identity
That is

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


From the above identity here
a=3x
b=y
c=z

expand according to identity ,we get

(3x+y+z)(9x^2+y^2+z^2-3xy-yz-yz-3.3x.y.z)

(3x+y+z)(9x^2+y^2+z^2-3xy-3xz-3xz)+3.3x.y.z-9xyz

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

.9xyz and -9xyz both hv oppste sign so become zero


Hope you understood

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