Question

Question 6
Multiply $(x^2+4y^2+z^2+2xy+xz–2yz)by(-z+x–2y.)$

Answer 1

$(x^2+4y^2+z^2+2xy+xz–2yz)(-z+x–2y)$
$=x^2(-z+x–2y)+4y^2(-z+x–2y)+z^2(-z+x-2y)+2xy(-z+x-2y)+xz(-z+x-2y)-2yz(-z+x-2y)$
$=-x^{2}z+x^3–2x^{2}y–4y^{2}z+4x{y}^2–8y^3–z^3+x{z}^2-2y{z}^2-2xyz+2x^{2}y-4x{y}^2-x{z}^2+x^{2}z-2xyz+2y{z}^2-2xyz+4y^{2}z$
$=(-x^{2}z+x^{2}z)+x^3+(-2x^{2}y+2x^{2}y)+(-4y^{2}z+4y^{2}z)+(4x{y}^2-4x{y}^2)-8y^3-z^3+(x{z}^2-x{z}^2)+(-2y{z}^2+2y{z}^2)+(-2xyz-2xyz-2xyz)$
$=x^3-8y^3-z^3-6xyz$

Alternate method:

$(x–2y–z)(x^2+4y^2+z^2+2xy+xz-2yz)$
$=(x-2y-z)\left[\right(x{)}^2+(-2y{)}^2+(-z{)}^2-\left(x\right)(-2y)-(-2y)(-z)-\left(x\right)(-z)]$
$=(x{)}^3+(-2y{)}^3+(-z{)}^3–3(x\left)\right(-2y\left)\right(-z)$
$[Usingtheidentity,a^3+b^3+c^3-3abc=(a+b+c\left)\right(a^2+b^2+c^2–ab–bc–ca\left)\right]$
$=x^3–8y^3–z^3–6xyz$