Question

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

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

Answer 1

Given, $(4y^2-9x^2)(16y^4+36x^2{y}^2+81x^4)$

Using Identity,

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Comparing given equation with identity we get,

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

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

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

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

Substituting $x=-2,y=1$ we get,

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

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

$=64-729\times 64$

$=64(1-729)$

$=64\times -728$

$=-46592$