Question

If $x=3$ and $y=-1,$ find the values of the following using identity:

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

Answer 1

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

Using Identity:

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

Comparing given equation with identity we get,

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

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

$729y^6-64x^6=(9y^2-4x^2)(81y^4+36x^2{y}^2+16x^4)$

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

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

Substituting $x=3$ and $y=-1$ we get.

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

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

$=729(1-64)$

$=729x-63$

$=-45927$

$\therefore (9y^2-4x^2)(81y^4+36x^2{y}^2+16x^4)=-45927$
$\mathrm{When}x=3\mathrm{and}y=-1$