(i) (99)3 [Using identity (a−b)3=a3−b3−3ab(a−b)]⇒(100−1)3=(100)3+(−1)3+3(100)(−1)(100−1)=1000000−1−300(100−1)=1000000−1−30000+300=970299
(ii) (102)3 [Using identity (a+b)3=a3+b3+3ab(a+b)]⇒(100+2)3=(100)3+23+3(100)(2)(100+2)=1000000+8+600(100+2)=1000000+8+60000+1200=1061208
(iii) (998)3 [Using identity (a−b)3=a3−b3−3ab(a−b)]⇒(1000−2)3=(1000)3+(−2)3+3(1000)(−2)(998)=(1000)3−8−6000(998)=1000000000−8−5988000=994011992