(i) 1033=(100+3)3
=(100)3+(3)3+3×100×3(100+3) [Using the identity, (a+b)3=a3+b3+3ab(a+b)]
=1000000+27+900(103)
=1000027+92700=1092727
(ii) 101×102=(100+1)(100+2)
=(100)2+100(1+2)+1×2 [Using the identity, (x+a)(x+b)=x2+x(a+b)+ab]
=10000+300+2=10302
(iii) (999)2=(1000−1)2
=(1000)2+(1)2–2×1000×1 [Using the identity, (a–b)2=a2+b2–2ab]
=1000000+1–2000=998001