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Answer 1

I think question is
Q : Factorise
$1) 9 {(x + y)}^{2} - {x}^{2} \\ = > {(3(x + y))}^{2} - {x}^{2} \\ = > (3(x + y) - x)(3(x + y) + x) \\ = > (2x + 3y)(4x + 3y)$
2)
$9 {x}^{2} - 4 {(y + 2x)}^{2} \\ = > {(3x)}^{2} - {(2(y + 2x))}^{2} \\ = > (3x - 2y - 4x)(3x + 2y + 4x) \\ = > ( - x - 2y)(7x + 2y)$

Answer 2

( 1 ) . 9(x + y)² - x²

[3(x + y)]² - x²
we know the identity (a - b)(a + b) = (a² - b²)

now,use it here.

we get,
(3x + 3y - x)(3x + 3y + x)

(2x + 3y)(4x + 3y)

( 2 ) . 9x² - 4(y + 2x)²

(3x)² - [2(y + 2x)]²
we can see this is in the form of
(a² - b²)
so, you can use identity
(a - b)(a + b) = (a² - b²)

(3x - 2y - 4x)(3x + 2y + 4x)

(-x - 2y)(7x + 2y)