The Total solution for NCERT class 6-12
32a3 + 108b3= 4(8a3 + 27b3) = 4 [(2a)3 +(3 b)3]= 4(2a + 3b) [(2a)2 – 2a x 3b + (3b)2]= 4(2a + 3b) (4a2 – 6ab + 9b2)