Find the product :
(i) (2x + 3y) (2x - 3y) $(4 x^{2} + 9 y^{2}$ )

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Solution

(i) We have,
(2x + 3y) (2x - 3y) $(4 x^{2} + 9 y^{2}$ )
= {(2x+3y) (2x-3y)} $(4 x^{2} + 9 y^{2})$
= {(2x + 3y) (2x - 3y)} $(4 x^{2} + 9 y^{2})$
$= {{(2 x)}^{2} - {(3 y)}^{2}} (4 x^{2} + 9 y^{2})$ [Using : (a + b) (a - b) = $a^{2} - b^{2}$ ]
$= (4 x^{2} - 9 y^{2}) (4 x^{2} + 9 y^{2})$

$= {(4 x^{2})}^{2} - {(9 y^{2})}^{2}$ [Using : (a + b) (a - b) = $a^{2} - b^{2}$ ]

$= 16 x^{4} - 81 y^{4}$

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Similar questions

Find the product of $(2 x - 3 y)$ and $(4 x^{2} + 9 y^{2} + 6 x y)$ .

\frac{2 x}{3 y} = \frac{3 y}{4 z}

4 x^{2} + 9 y^{2} + 16 z^{2} = (2 x - 3 y + 4 z) (2 x + 3 y + 4 z)

\frac{2 x}{3 y} = \frac{3 y}{4 z}

4 x^{2} + 9 y^{2} + 16 z^{2} = (2 x - 3 y + 4 z) (2 x + 3 y + 4 z)

4 x^{2} + 9 y^{2} = 40, x y = 2

2 x + 3 y

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