Question

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

• A. $0.999$
• B. $0.998$
• C. $0.997$
• D. $0.995$

Answer 1

Answer: (C)

${\left(0.999\right)}^3={\left(1-0.001\right)}^3$

$={}^3{C}_0-{}^3{C}_1\left(0.001\right)+{}^3{C}_2{\left(0.001\right)}^2-{}^3{C}_3{\left(0.001\right)}^3$

$=1-0.003+3\left(0.000001\right)-\left(0.000000001\right)$

$=0.997$