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

Options

A. (a) `0.999`
B. (b) `0.998`
C. (c) `0.997`
D. (d) `0.995`

Answer

To find the value of \( (0.999)^3 \) using the Binomial Theorem, we can express \( 0.999 \) as \( 1 - 0.001 \). Thus, we can rewrite the expression as:

\[ (0.999)^3 = (1 - 0.001)^3 \]

According to the Binomial Theorem, we can expand \( (a - b)^n \) as follows:

\[ (a - b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} (-b)^k \]

In our case, \( a = 1 \), \( b = 0.001 \), and \( n = 3 \). Therefore, we can expand \( (1 - 0.001)^3 \):

\[ (1 - 0.001)^3 = \sum_{k=0}^{3} \binom{3}{k} (1)^{3-k} (-0.001)^k \]

Now, we calculate each term in the expansion:

1. For \( k = 0 \):
\[ \binom{3}{0} (1)^{3} (-0.001)^0 = 1 \cdot 1 \cdot 1 = 1 \]

2. For \( k = 1 \):
\[ \binom{3}{1} (1)^{2} (-0.001)^1 = 3 \cdot 1 \cdot (-0.001) = -0.003 \]

3. For \( k = 2 \):
\[ \binom{3}{2} (1)^{1} (-0.001)^2 = 3 \cdot 1 \cdot 0.000001 = 0.000003 \]

4. For \( k = 3 \):
\[ \binom{3}{3} (1)^{0} (-0.001)^3 = 1 \cdot 1 \cdot (-0.000000001) = -0.000000001 \]

Now, we combine all the terms:

\[ (1 - 0.001)^3 = 1 - 0.003 + 0.000003 - 0.000000001 \]

Calculating this step by step:

- Start with \( 1 \)
- Subtract \( 0.003 \):
\[ 1 - 0.003 = 0.997 \]
- Add \( 0.000003 \):
\[ 0.997 + 0.000003 = 0.997003 \]
- Subtract \( 0.000000001 \):
\[ 0.997003 - 0.000000001 \approx 0.997003 \]

Now, rounding \( 0.997003 \) to three decimal places gives us:

\[ 0.997 \]

Thus, the value of \( (0.999)^3 \) correct to three decimal places is:

\[ \boxed{0.997} \]