(2a+3b)3=(2a)3+(3b)3+3(2a)(3b)(2a+3b)
Using, (a+b)3=a3+b3+3ab(a+b)
=8a3+27b3+18ab(2a+3b)
=8a3+27b3+36a2b+54ab2 .........(i)
(2a−3b)3=(2a)3−(3b)3−3(2a)(3b)(2a−3b)
Using, (a−b)3=a3−b3−3ab(a−b)
=8a3−27b3−18ab(2a−3b)
=8a3−27b3−36a2b+54ab2......(ii)
Subtracting (i) from (ii) we get,
(2a+3b)3−(2a−3b)3=8a3+27b3+36a2b+54ab2−(8a3−27b3−36a2b+54ab2)
=8a3+27b3+36a2b+54ab2−8a3+27b3+36a2b−54ab2
=54b3+72a2b