Question

(2x+3y)square2 +(2x-3y)square2

Answer 1

$\blue{\huge{\boxed{\star\: Solution \: \star}}} Firstly, \boxed{open \: brackets \: by \: identity}$

$(a + b) ^{2} = a ^{2} + {b}^{2} + 2ab$

$(a - b) {}^{2} = {a}^{2} + {b}^{2} - 2ab$

So,(2x+3y)^2 + (2x-3y)^2

===>
$(2x) {}^{2} + (3y) {}^{2} + 2(2x)(3y) + (2x) {}^{2} + (3y) {}^{2} - 2(2x)(3y)$

===>
$4x {}^{2} + 9y {}^{2} + 12xy + 4x {}^{2} + 9y {}^{2} - 12xy$

So,

$\green{ \boxed{8x {}^{2} + 18y {}^{2} }}$

Answer 2

❤❤here is your answer ✌ ✌
$= (2x + 3y) {}^{2} +( 2x - 3y) {}^{2} \\ = 4 {x}^{2} + 9 {y}^{2} + 12xy \\ + 4 {x}^{2} + 9 {y}^{2} - 12xy \\ = 8 {x}^{2} + 18 {y}^{2}$
I hope it helps you