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