If $x = -2$ and $y = 1$, by using an identity find the value of the following:
$ \left(4 y^{2}-9 x^{2}\right)\left(16 y^{4}+36 x^{2} y^{2}+81 x^{4}\right) $

Given: 

$x = -2$ and $y = 1$

To do: 

We have to find the value of \( \left(4 y^{2}-9 x^{2}\right)\left(16 y^{4}+36 x^{2} y^{2}+81 x^{4}\right) \).

Solution: 

We know that,

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

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

Therefore,

$(4 y^{2}-9 x^{2})(16 y^{4}+36 x^{2} y^{2}+81 x^{4})=(4 y^{2}-9 x^{2})[(4 y^{2})^{2}+4 y^{2} \times 9 x^{2}+(9 x^{2})^{2}]$

$=(4 y^{2})^{3}-(9 x^{2})^{3}$

$=64 y^{6}-729 x^{6}$

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

$=64 \times 1-729 \times(64)$

$=64-46656$

$=-46592$

Hence, $(4 y^{2}-9 x^{2})(16 y^{4}+36 x^{2} y^{2}+81 x^{4})=-46592$.  

Tutorialspoint

Updated on: 10-Oct-2022

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