1.

Evaluate `(i) (102)^3` `(ii) (999)^3`

Answer» We have
`(i) (102)^3=(100+2)^3`
`=(100)^3+2^3+3xx 100 xx 2 xx (100 +2)`
`=1000000+8(600 xx 102)`.
`=1000008 +61200 =1061208`.
`(ii) (999)^3=(1000-1)^3`
`=(1000)^3-1^3-(3xx 1000xx 1)(1000-1)`
`=1000000000-1-(3000xx 999)`
`=999999999-2997000=997002999`.


Discussion

No Comment Found