(i) We have,
(2x+3y)2=(2x)2+2×(2x)×(3y)+(3y)2 [Using : (a+b)2=a2+2ab+b2]
= 4x2+12xy+9y2
(ii) We have,
(2x−3y)2=(2x)2−2×(2x)×(3y)+(3y)2 [Using : (a−b)2=a2−2ab+b2]
= 4x2−12xy+9y2
(iii) We have,
(2x+3y)(2x−3y)=(2x)2−(3y)2 [Using : (a+b)(a−b)=a2−b2]
4x2−9y2.