Using suitable identity, evaluate the following

(i) $103^{3}=(100+3)^{3}$

$=(100)^{3}+(3)^{3}+3 \times 100 \times 3 \times(100+3)$ [using identity, $\left.(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)\right]$

$=1000000+27+900(103)$

$=1000027+92700=1092727$

(ii) $101 \times 102=(100+1)(100+2)$

$=(100)^{2}+100(1+2)+1 \times 2 \quad$ [using identity, $\left.(x+a)(x+b)=x^{2}+x(a+b)+a b\right]$

$=10000+300+2=10302 .$

(iii) $(999)^{2}=(1000-1)^{2}$

$=(1000)^{2}+(1)^{2}-2 \times 1000 \times 1$              [using identity, $(a-b)^{2}=a^{2}+b^{2}-2 a b$ ]

$=1000000+1-2000=998001$