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