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Question

Question 25
Using suitable identity, evaluate the following
(i) 1033

(ii) 101×102

(iii) 9992

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Solution

(i) 1033=(100+3)3
=(100)3+(3)3+3×100×3(100+3) [Using the identity, (a+b)3=a3+b3+3ab(a+b)]
=1000000+27+900(103)
=1000027+92700=1092727

(ii) 101×102=(100+1)(100+2)
=(100)2+100(1+2)+1×2 [Using the identity, (x+a)(x+b)=x2+x(a+b)+ab]
=10000+300+2=10302

(iii) (999)2=(10001)2
=(1000)2+(1)22×1000×1 [Using the identity, (ab)2=a2+b22ab]
=1000000+12000=998001

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