If $x = 3$ and $y = -1$, find the values of each of the following using in identity:
$ \left(9 y^{2}-4 x^{2}\right)\left(81 y^{4}+36 x^{2} y^{2}+16 x^{4}\right) $

Given: 

$x = 3$ and $y = -1$

To do: 

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

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,

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

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

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

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

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

$=729-46656$

$=-45927$

Hence, $(9 y^{2}-4 x^{2})(81 y^{4}+36 x^{2} y^{2}+16 x^{4})=-45927$.

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Updated on: 10-Oct-2022

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