The Total solution for NCERT class 6-12
a3 + 3a2b + 3ab2 + b3 –8= (a + b)3 – (2)3= (a + b -2)[(a + b)2 + (a +b)x2 + (2)2]= (a + b-2) (a2 + b2 + 2ab + 2a + 2b + 4)= (a + b – 2) (a2 + b2 + 2ab + 2(a + b) + 4]= (a + b – 2) [(a + b)2 + 2(a + b) + 4}