Question

Question 35
Find the following product:
$(2x-y+3z)(4x^2+y^2+9z^2+2xy+3yz-6xz)$

Answer 1

$(2x-y+3z)(4x^2+y^2+9z^2+2xy+3yz-6xz)$
$2x(4x^2+y^2+9z^2+2xy+3yz-6xz)-y(4x^2+y^2+9z^2+2xy+3yz-6xz)+3z(4x^2+y^2+9z^2+2xy+3yz-6xz)$
$=8x^3+2x{y}^2+18x{z}^2+4x^{2}y+6xyz-12x^{2}z-4x^{2}y-y^3-9y{z}^2-2x{y}^2$
$-3y^{2}z+6xyz+12x^{2}z+3y^{2}z+27z^3+6xyz+9y{z}^2-18x{z}^2$
$=8x^3+(2x{y}^2-2x{y}^2)+(18x{z}^2-18x{z}^2)+(4x^{2}y-4x^{2}y)+(6xyz+6xyz+6xyz)$
$+(-12x^{2}z+12x^{2}z)-y^3+(-9y{z}^2+9y{z}^2)+(-3y^{2}z+3y^{2}z)+27z^3$
$=8x^3+18xyz-y^3+27z^3$
$=8x^3-y^3+27z^3+18xyz$

Alternate method:

$(2x-y+3z)(4x^2+y^2+9z^2+2xy+3yz-6xz)$
$=(2x-y+3z)(2x{)}^2+(-y{)}^2+(3z{)}^2-(2x\left)\right(-y)-(-y\left)\right(3z)-(2x\left)\right(3z)$
$=(2x{)}^3+(-y{)}^3+(3z{)}^3-3(2x\left)\right(-y\left)\right(3z)$
$[Usingtheidentity,(a+b+c\left)\right(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc]$
$=8x^3-y^3+27z^3+18xyz$