Question

Evaluate `(i) 101xx 102`
`(ii) (999)^2`
`(iii) (997)^2`

Answer 1

We have
`(i) 101xx 102= 101 xx (100+2)`
`=(101 xx 100) +(101 x2)=10100 +202 =10302`.
`(ii) (999)^2=(1000-1)^2`
`=(1000)^2+1^2-(2xx 1000xx1)`
`=1000000+1-2000`
`=(1000001-2000) =998001`.
`(iii) (997)^2=1(1000-3)^2`
`=(1000)^2+3^2-(2xx1000xx 3)`
`=1000000+9-6000=994009`.