Given,
(2x+3y)3−(2x−3y)3.
This is of the form (a3−b3)=(a−b)(a2+ab+b2), [Here a = (2x+3y)3, b = (2x−3y)3]
[1]
Applying the identity,
(2x+3y)3−(2x−3y)3
=[(2x+3y)−(2x−3y)][(2x+3y)2+(2x+3y)(2x−3y)+(2x−3y)2]
[1]
=[2x+3y−2x+3y][4x2+12xy+9y2+4x2−9y2+4x2−12xy+9y2]
=6y(12x2+9y2)
=72x2y+54y3
[1]