If x = – 2 and y = 1, by using an identity find the value of the following:

Solution:

Given,

(a) $\left(4 y^{2}-9 x^{2}\right)\left(16 y^{4}+36 x^{2} y^{2}+81 x^{4}\right)$

We know that,

$a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\left(4 y^{2}-9 x^{2}\right)\left(16 y^{4}+36 x^{2} y^{2}+81 x^{4}\right)$

can be written as,

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

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

$\Rightarrow 64 y^{6}-729 x^{6} \ldots 1$

Substitute $x=-2$ and $y=1$ in eq 1

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

$\Rightarrow 64(1)^{6}-729(-2)^{6}$

$\Rightarrow 64-729(64)$

$\Rightarrow 64(1-729)$

$\Rightarrow 64(-728)$

$\Rightarrow-46592$

Hence, the value of $\left(4 y^{2}-9 x^{2}\right)\left(16 y^{4}+36 x^{2} y^{2}+81 x^{4}\right)$ is $-46592$

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

We know that,

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

can be witten as,

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

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

$\Rightarrow\left(8 / x^{3}\right)-\left(x^{3} / 8\right) \ldots 1$

Substitute x = -2 in eq 1

$\Rightarrow\left(8 /(-2)^{3}\right)-\left((-2)^{3} / 8\right)$

$\Rightarrow(8 /-8)-(-8 / 8)$

$\Rightarrow-1+1$

$\Rightarrow 0$

Hence, the value of $(2 / x-x / 2)\left(4 / x^{2}+x^{2} / 4+1\right)$ is 0

(c) $(5 y+15 / y)\left(25 y^{2}-75+225 / y^{2}\right)$

We know that,

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

$(5 y+15 / y)\left(25 y^{2}-75+225 / y^{2}\right)$

can be written as,

$\Rightarrow(5 y+15 / y)\left[(5 y)^{2}+(15 y)^{2}-(5 y)(15 / y)\right]$

$\Rightarrow(5 y)^{3}+(15 / y)^{3}$

$\Rightarrow 125 y^{3}+\left(3375 / y^{3}\right) \ldots 1$

Substitute y = 1 in eq 1

$\Rightarrow 125(1)^{3}+\left(3375 /(1)^{2}\right)$

$\Rightarrow 125+3375$

$\Rightarrow 3500$

Hence, the value of $(5 y+15 / y)\left(25 y^{2}-75+225 / y^{2}\right)$ is 3500 .