Find x = 3 and y = -1, Find the values of each of the following using in identity:

Solution:

(a)  We know that,

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

$\left(9 x^{2}-4 x^{2}\right)\left(81 y^{4}+36 x^{2} y^{2}+16 x^{4}\right)$ can be written as,

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

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

$\Rightarrow 729 y^{6}-64 x^{6}$

Substitute the value $x=3, y=-1$ in $729 y^{6}-64 x^{6}$ we get,

$\Rightarrow 729 y^{6}-64 x^{6}$

$\Rightarrow 729(-1)^{6}-64(3)^{6}$

$\Rightarrow 729(1)-64(729)$

$\Rightarrow 729-46656$

$\Rightarrow-45927$

Hence, the product value of $\left(9 x^{2}-4 x^{2}\right)\left(81 y^{4}+36 x^{2} y^{2}+16 x^{4}\right)=-45927$

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

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

$\Rightarrow\left(27 / x^{3}\right)-\left(x^{3} / 27\right) \ldots 1$

Substitute $x=3$ in eq 1

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

$\Rightarrow(27 / 27)-(27 / 27)$

$\Rightarrow 0$

Hence, the value of $(3 / x-5 / y)\left(9 / x^{2}+25 / y^{2}+15 / x y\right)$ is 0

(c) Given,

$(x / 7+y / 3)\left(x^{2} / 49+y^{2} / 9-x y / 21\right)$

 We know that,

$a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)(x / 7+y / 3)\left(x^{2} / 49+y / 29-x y / 21\right)$

Can be written as,

$\Rightarrow(x / 7+y / 3)\left[(x / 7)^{2}+(y / 3)^{2}-(x / 7)(y / 3)\right]$

$\Rightarrow(x / 7)^{3}+(y / 3)^{3}$

$\Rightarrow\left(x^{3} / 343\right)+\left(y^{3} / 27\right) \ldots 1$

Substitute x = 3, y = -1 in eq 1

$\Rightarrow\left(3^{3} / 343\right)+\left((-1)^{3} / 27\right)$

$\Rightarrow(27 / 343)-(1 / 27)$

Taking least common multiple, we get

$\Rightarrow \frac{27 * 27}{343 * 27}-\frac{1 * 343}{27 * 343}$

$\Rightarrow \frac{729}{9261}-\frac{343}{9261}$

$\Rightarrow \frac{729-343}{9261}$

$\Rightarrow \frac{386}{9261}$

Hence, the value of $(x / 7+y / 3)\left(x^{2} / 49+y^{2} / 9-x y / 21\right)$

= 386/9261 

(d) Given, 

$(x / 4-y / 3)\left(x^{2} / 16+y^{2} / 9-+x y / 21\right)$

We know that,

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

$\left(x^{4}-y^{3}\right)\left(x^{2} / 16+y^{2} / 9+x y / 21\right)$

Can be written as,

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

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

$\Rightarrow\left(x^{3} / 64\right)-\left(y^{3} / 27\right) \ldots .1$

Substitute x = 3, y = -1 in eq 1

$\Rightarrow\left(3^{3} / 343\right)-\left((-1)^{3} / 27\right)$

$\Rightarrow(27 / 64)+(1 / 27)$

Taking least common multiple, we get

$\Rightarrow \frac{27 * 27}{64 * 27}+\frac{1 * 64}{27 * 64}$

$\Rightarrow 729 / 1728+64 / 1728$

$\Rightarrow \frac{729+64}{1728}$

$\Rightarrow 793 / 9261$

Hence, the value of $(x / 4-y / 3)\left(x^{2} / 16+y^{2} / 9+x y / 21\right)$

= 793/1728

(e) Given,

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

We know that,

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

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

can be written as, $\Rightarrow(5 / x+5 x)\left[(5 / x)^{2}+(5 x)^{2}-(5 / x)(5 x)\right]$

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

$\Rightarrow 125 / x^{3}+125 x^{3} \ldots 1$

Substitute x = 3, in eq 1

$\Rightarrow 125 / 3^{3}+125(3)^{3}$

$\Rightarrow 125 / 27+125 * 27$

$\Rightarrow 125 / 27+3375$

Taking least common multiple, we get

$\Rightarrow \frac{125}{27}+\frac{3375 * 27}{27 * 1}$

$\Rightarrow 125 / 27+91125 / 27$

$\Rightarrow \frac{125+91125}{27}$

$\Rightarrow 91250 / 25$

Hence, the value of $(5 / x+5 x)\left(25 / x^{2}-25+25 x^{2}\right)$ is $91250 / 25$