Question
Factorize the following expressions:
$x^3-8 y^3+27 z^3+18 x y z$
 

Answer

$x^3-8 y^3+27 z^3+18 x y z$
$=x^3+(-2 y)^3+(3 z) 3-3 x x \times(-2 y)(3 z)$
$=(x+(-2 y)+3 z)\left(x^2+(-2 y)^2+(3 z)^2-x(-2 y)-(-2 y)(3 z)-3 z(x)\right)$
${\left[\because a^3+b^3+c^3-3 a b c=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)\right]}$
$=(x-2 y+3 z)\left(x^2+4 y^2+9 z^2+2 x y+6 y z-3 z x\right)$
$\therefore x^3-8 y^3+27 z^3+18 x y z=(x-2 y+3 z)\left(x^2+4 y^2+9 z^2+2 x y+6 y z-3 z x\right)$

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