# Prove the following

Question:

Fi nd (2x – y + 3z) (4x2 + y2 + 9z+ 2xy + 3yz – 6xz).

Solution:

$(2 x-y+3 z)\left(4 x^{2}+y^{2}+9 z^{2}+2 x y+3 y z-6 x z\right)$

$=2 x\left(4 x^{2}+y^{2}+9 z^{2}+2 x y+3 y z-6 x z\right)-y\left(4 x^{2}+y^{2}+9 z^{2}+2 x y+3 y z-6 x z\right)$

$+3 z\left(4 x^{2}+y^{2}+9 z^{2}+2 x y+3 y z-6 x z\right)$

$=8 x^{3}+2 x y^{2}+18 x z^{2}+4 x^{2} y+6 x y z-12 x^{2} z-4 x^{2} y-y^{3}-9 y z^{2}-2 x y^{2}$

$-3 y^{2} z+6 x y z+12 x^{2} z+3 y^{2} z+27 z^{3}+6 x y z+9 y z^{2}-18 x z^{2}$

$=8 x^{3}+\left(2 x y^{2}-2 x y^{2}\right)+\left(18 x z^{2}-18 x z^{2}\right)+\left(4 x^{2} y-4 x^{2} y\right)+(6 x y z+6 x y z+6 x y z)$

$+\left(-12 x^{2} z+12 x^{2} z\right)-y^{3}+\left(-9 y z^{2}+9 y z^{2}\right)+\left(-3 y^{2} z+3 y^{2} z\right)+27 z^{3}$

$=8 x^{3}+18 x y z-y^{3}+27 z^{3}$

$=8 x^{3}-y^{3}+27 z^{3}+18 x y z$

Alternate Method

$(2 x-y+3 z)\left(4 x^{2}+y^{2}+9 z^{2}+2 x y+3 y z-6 x z\right)$

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

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

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

$=8 x^{3}-y^{3}+27 z^{3}+18 x y z$