Question
Multiply:
$\left(x^2+4 y^2+z^2+2 x y+x z-2 y z\right)$ by $(x-2 y-z)$
$\left(x^2+4 y^2+z^2+2 x y+x z-2 y z\right)$ by $(x-2 y-z)$
✓
Answer
$=(x-2 y-z)\left(x^2+4 y^2+z^2+2 x y+x z-2 y z\right)$
$=(x+(-2 y)+(-z))\left(x^2+(-2 y)^2+(-z)^2-x(-2 y)-(-2 y)(-z)-(-z) x\right)$
$=x^3+(-2 y)^3+(-z)^3-3 \times x(-2 y)(-z)\left[\because(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)=a^3+b^3+c^3-3 a b c\right]$
$=x^3-8 y^3-z^3+3 \times x \times 2 y z$
$=x^3-8 y^3-z^3-6 x y z$
$=(x+(-2 y)+(-z))\left(x^2+(-2 y)^2+(-z)^2-x(-2 y)-(-2 y)(-z)-(-z) x\right)$
$=x^3+(-2 y)^3+(-z)^3-3 \times x(-2 y)(-z)\left[\because(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)=a^3+b^3+c^3-3 a b c\right]$
$=x^3-8 y^3-z^3+3 \times x \times 2 y z$
$=x^3-8 y^3-z^3-6 x y z$
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