Using suitable identity , evaluate the following
(i) `103^(3)` (ii) `101xx102`(iii) `999^(2)`

Explanation

(i) `103^(3)=(100+3)^(3)`
`=(100)^(3)(3)^(3)+3xx100xx3xx(100+3)` [using identity, `(a+b)^(3)=a^(3)+b^(3)+3ab(a+b)`]
`=10000000+27+900(103)`
`=1000027+92700=1092727`
(ii) `101xx102=(100+1)(100+2)`
`=(100)^(2)+100(1+2)+1xx2` [using identity `(x+a)(x+b)=x^(2)+x(a+b)+ab`]
`=10000+300+2=10302`
(iii) `(999)^(2)=(1000-1)^(2)`
`=(1000)^(2)+(1)^(2)-2zz1000xx1` [using identity ,`(a-b)^(2)=a^(2)+b^(2)-2ab`]
`=1000000+1-2000=998001`