Find the following products

Solution:

(a) Given,

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

We know that,

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

can we written as

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

$\Rightarrow(3 x)^{3}+(2 y)^{3}$

$\Rightarrow 27 x^{3}+8 y^{3}$

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

$=27 x^{3}+8 y^{3}(b)(4 x-5 y)\left(16 x^{2}+20 x y+25 y^{2}\right)$

(b) Given,

$(4 x-5 y)\left(16 x^{2}+20 x y+25 y^{2}\right)$

We know that,

$a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)(4 x-5 y)\left(16 x^{2}+20 x y+25 y^{2}\right)$

can we written as

$\left.\Rightarrow(4 x-5 y)\left[(4 x)^{2}+(4 x)(5 y)+(5 y)^{2}\right)\right]$

$\Rightarrow(4 x)^{3}-(5 y)^{3}$

$\Rightarrow 16 x^{3}-25 y^{3}$

Hence, the value of $(4 x-5 y)\left(16 x^{2}+20 x y+25 y^{2}\right)$

$=16 x^{3}-25 y^{3}$

(c) Given

$\left(7 p^{4}+q\right)\left(49 p^{8}-7 p^{4} q+q^{2}\right)$

We know that,

$a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\left(7 p^{4}+q\right)\left(49 p^{8}-7 p^{4} q+q^{2}\right)$

can be written as

$\left.\Rightarrow\left(7 p^{4}+q\right)\left[\left(7 p^{4}\right)^{2}-\left(7 p^{4}\right)(q)+(q)^{2}\right)\right]$

$\Rightarrow\left(7 p^{4}\right)^{3}+(q)^{3}$

$\Rightarrow 343 p^{12}+q^{3}$

(d) Given,

$(x / 2+2 y)\left(x^{2} / 4-x y+4 y^{2}\right)$

We know that,

$a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)(x / 2+2 y)\left(x^{2} / 4-x y+4 y^{2}\right)$

can be written as

$\Rightarrow(x / 2+2 y)\left[(x / 2)^{2}-x / 2(2 y)+(2 y)^{2}\right]$

$\Rightarrow(x / 2)^{3}+(2 y)^{3}$

$\Rightarrow x^{3} / 8+8 y^{3}$

(e) Given, 

$(3 / x-5 / y)\left(9 / x^{2}+25 / y^{2}+15 / x y\right)$

We know that,

$a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)(3 / x-5 / y)\left(9 / x^{2}+25 / y^{2}+15 / x y\right)$

Can be written as,

$\Rightarrow(3 / x-5 / y)(3 / x)^{2}+(5 / y)^{2}+(3 / x)(5 / y)$

$\Rightarrow(3 / x)^{3}-(5 / y)^{3}$

$\Rightarrow\left(27 / x^{3}\right)-\left(125 / y^{3}\right)$

Hence, the value of $(3 / x-5 / y)\left(9 / x^{2}+25 / y^{2}+15 / x y\right)$

$=\left(27 / x^{3}\right)-\left(125 / y^{3}\right)$

(f)  Given, 

$(3+5 / x)\left(9-15 / x+25 / x^{2}\right)$

We know that,

$a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)(3+5 / x)\left(9-15 / x+25 / x^{2}\right)$

can be written as,

$\Rightarrow(3+5 / x)\left[\left(3^{2}\right)-3(5 / x)+(5 / x)^{2}\right]$

$\Rightarrow(3)^{3}+(5 / x)^{3}$

$\Rightarrow 27+125 / x^{3}$

Hence, the value of $(3+5 x)\left(9-15 / x+25 / x^{2}\right)$ is $27+125 / x^{3}$

(g) Given, 

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

We know that,

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

can be written as,

$\Rightarrow(2 / x+3 x)\left[(2 / x)^{2}+(3 x)^{2}-(2 / x)(3 x)\right]$

$\Rightarrow(2 / x)^{3}+(3 x)^{3}$

$\Rightarrow 8 / x^{3}+9 x^{3}$

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

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

Given, 

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

We know that,

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

can be written as,

$\Rightarrow\left(3 / x-2 x^{2}\right)\left[(3 / x)^{2}+\left(2 x^{2}\right)^{2}-(3 / x)\left(2 x^{2}\right)\right]$

$\Rightarrow\left(3 / x-2 x^{2}\right)\left[\left(9 / x^{2}\right)+4 x^{4}-(3 / x)\left(2 x^{2}\right)\right]$

$\Rightarrow(3 / x)^{3}-\left(2 x^{2}\right)^{3}$

$\Rightarrow 27 / x^{3}-8 x^{6}$

Hence, $\left(3 / x-2 x^{2}\right)\left(9 / x^{2}+4 x^{4}-6 x\right)$ is $27 / x^{3}-8 x^{6}$

(i) $(1-x)\left(1+x+x^{2}\right)$

Sol:

Given, $(1-x)\left(1+x+x^{2}\right)$

We know that, $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)(1-x)\left(1+x+x^{2}\right)$

can be written as, $\Rightarrow(1-x)\left[\left(1^{2}+(1)(x)+x^{2}\right)\right]$

$\Rightarrow(1)^{3}-(x)^{3}$

$\Rightarrow 1-x^{3}$

Hence, the value of $(1-x)\left(1+x+x^{2}\right)$ is $1-x^{3}$

(j) Given,

$(1+x)\left(1-x+x^{2}\right)$

We know that,

$a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)(1+x)\left(1-x+x^{2}\right)$

$a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)(1+x)\left(1-x+x^{2}\right)$

can be written as,

$\Rightarrow(1+x)\left[\left(1^{2}-(1)(x)+x^{2}\right)\right]$

$\Rightarrow(1)^{3}+(x)^{3}$

$\Rightarrow 1+x^{3}$

Hence, the value of $(1+x)\left(1+x-x^{2}\right)$ is $1+x^{3}(k)\left(x^{2}-1\right)\left(x^{4}+x^{2}+1\right)$

(k) Given, 

$\left(x^{2}-1\right)\left(x^{4}+x^{2}+1\right)$

We know that,

$a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\left(x^{2}-1\right)\left(x^{4}+x^{2}+1\right)$

can be written as,

$\Rightarrow\left(x^{2}-1\right)\left[\left(x^{2}\right)^{2}-1^{2}+\left(x^{2}\right)(1)\right]$

$\Rightarrow\left(x^{2}\right)^{3}-1^{3}$

$\Rightarrow x^{6}-1$

Hence, $\left(x^{2}-1\right)\left(x^{4}+x^{2}+1\right)$ is $x^{6}-1$

(l) Given, 

$\left(x^{2}+1\right)\left(x^{6}-x^{3}+1\right)$

We know that, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)\left(x^{2}+1\right)\left(x^{6}-x^{3}+1\right)$

can be written as, $\Rightarrow\left(x^{3}+1\right)\left[\left(x^{3}\right)^{2}-\left(x^{3}\right)(1)+1^{2}\right]$

$\Rightarrow\left(x^{3}\right)^{3}+1^{3}$

$\Rightarrow x^{9}+1$

Hence, the value of $\left(x^{2}+1\right)\left(x^{6}-x^{3}+1\right)$ is $x^{9}+1$