Using binomial theorem, the value of ${(0.999)}^{3}$ correct to $3$ decimal places is

0.999

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0.998

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0.997

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0.995

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Solution

The correct option is C $0.997$
${(0.999)}^{3} = {(1 - 0.001)}^{3}$
$=^{3} C_{0} -^{3} C_{1} (0.001) +^{3} C_{2} {(0.001)}^{2} -^{3} C_{3} {(0.001)}^{3}$
$= 1 - 0.003 + 3 (0.000001) - (0.000000001)$
$= 0.997$

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