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Evaluate 1. (102)3 2. 140×96 3. (X+4) (x+10)

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Question 1. 102x3=306 2. 140x96=13440 3. (x+4)(x+10) =x+x of 4+10 = 2x of 14 = 28x answer.
Let us rewrite\xa0(102)3\xa0as\xa0(100+2)3Now using the identity\xa0(a+b)3=a3+b3+3ab(a+b), we get(100+2)3=1003+23+[(3×100×2)(100+2)]=1000000+8+(600×102)=1000000+8+61200=1061208Hence,\xa0(102)3=1061208


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