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EXERCISE 4.2 · Q27

Q.Using binomial theorem, find the value of (9.9)3(9.9)^3.

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$(9.9)^3=(10-0.1)^3=1000-3(100)(0.1)+3(10)(0.01)-(0.001)=1000-30+0.3-0.001=970.2 …

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