Prove the following

Now, $\left(x^{2}+4 y^{2}+z^{2}+2 x y+x z-2 y z\right)(-z+x-2 y)$

$=x^{2}(-z+x-2 y)+4 y^{2}(-z+x-2 y)+z^{2}(-z+x-2 y)+2 x y(-z+x-2 y)$

$+x z(-z+x-2 y)-2 y z(-z+x-2 y)$

$=-x^{2} z+x^{3}-2 x^{2} y-4 y^{2} z+4 x y^{2}-8 y^{3}-z^{3}+x z^{2}-2 y z^{2}-2 x y z+2 x^{2} y-4 x y^{2}$

$-x z^{2}+x^{2} z-2 x y z+2 y z^{2}-2 x y z+4 y^{2} z$

$=\left(-x^{2} z+x^{2} z\right)+x^{3}+\left(-2 x^{2} y+2 x^{2} y\right)+\left(-4 y^{2} z+4 y^{2} z\right)+\left(4 x y^{2}-4 x y^{2}\right)-8 y^{3}$

$-z^{3}+\left(x z^{2}-x z^{2}\right)+\left(-2 y z^{2}+2 y z^{2}\right)+(-2 x y z-2 x y z-2 x y z)$

$=x^{3}-8 y^{3}-z^{3}-6 x y z$

Alternate Method

Now, $(x-2 y-z)\left(x^{2}+4 y^{2}+z^{2}+2 x y+x z-2 y z\right)$

$=(x-2 y-z)\left[(x)^{2}+(-2 y)^{2}+(-z)^{2}-(x)(-2 y)-(-2 y)(-z)-(x)(-z)\right]$

$=(x)^{3}+(-2 y)^{3}+(-z)^{3}-3(x)(-2 y)(-z)$

[using identity, $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$ ]

$=x^{3}-8 y^{3}-z^{3}-6 x y z$