Find the following products:

Solution:

Given,

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

we know that,

$x^{3}+y^{3}+z^{3}-3 x y z$

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

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

$=27 x^{3}+8 y^{3}+8 z^{3}-36 x y z$

Hence, the value of $(3 x+2 y+2 z)\left(9 x^{2}+4 y^{2}+4 z^{2}-6 x y-4 y z-6 z x\right)$ is $27 x^{3}+8 y^{3}+8 z^{3}-36 x y z$

(b) $(4 x-3 y+2 z)\left(16 x^{2}+9 y^{2}+4 z^{2}+12 x y+6 y z-8 z x\right)$

we know that,

$x^{3}+y^{3}+z^{3}-3 x y z$

$=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$ so, $(4 x-3 y+2 z)\left(16 x^{2}+9 y^{2}+4 z^{2}+12 x y+6 y z-8 z x\right)$

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

$=64 x^{3}-27 y^{3}+8 z^{3}+72 x y z$

Hence, the value of $(4 x-3 y+2 z)\left(16 x^{2}+9 y^{2}+4 z^{2}+12 x y+6 y z-8 z x\right)$ is $64 x^{3}-27 y^{3}+8 z^{3}+72 x y z$

(c) $(2 a-3 b-2 c)\left(4 a^{2}+9 b^{2}+4 c^{2}+6 a b-6 b c+4 c a\right)$

we know that,

$x^{3}+y^{3}+z^{3}-3 x y z$

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

$=(2 a)^{3}+(-3 b)^{3}+(-2 c)^{3}-3(2 a)(-3 b)(-2 c)$

$=8 a^{3}-27 b^{3}-8 c^{3}-36 a b c$

Hence, the value of

(c) $(2 a-3 b-2 c)\left(4 a^{2}+9 b^{2}+4 c^{2}+6 a b-6 b c+4 c a\right)$ is $8 a^{3}-27 b^{3}-8 c^{3}-36 a b c$

(d) $(3 x-4 y+5 z)\left(9 x^{2}+16 y^{2}+25 z^{2}+12 x y-15 z x+20 y z\right)$

we know that, $x^{3}+y^{3}+z^{3}-3 x y z$

$=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$ so, $(3 x-4 y+5 z)\left(9 x^{2}+16 y^{2}+25 z^{2}+12 x y-15 z x+20 y z\right)$

$=(3 x)^{3}+(-4 y)^{3}+(5 z)^{3}-3(3 x)(-4 y)(5 z)=27 x^{3}-64 y^{3}+125 z^{3}+180 x y z$

Hence, the value of $(3 x-4 y+5 z)\left(9 x^{2}+16 y^{2}+25 z^{2}+12 x y-15 z x+20 y z\right)$ is $27 x^{3}-64 y^{3}+125 z^{3}+180 x y z$