\begin{tabular}{l} Use the Binomial Theorem to expand $$ (\mathrm{b}-2)^{3} $$ \ $$ \begin{array}{l}5 b^{3}-20 b^{2}+40 b-40 \ b^{3}-6 b^{2}+12 b-8 \ b^{3}-2 b^{2}+4 b-8 \ 4 b^{3}-12 b^{2}+16 b-8\end{array} $$ \ \hline\end{tabular}

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UpStudy AI · Accepted Answer

To expand $$ (\mathrm{b}-2)^{3} $$ using the Binomial Theorem, we can apply the formula:

$$
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
$$

In this case, we have $$ a = b $$, $$ b = -2 $$, and $$ n = 3 $$. Thus, we can write:

$$
(b - 2)^3 = \sum_{k=0}^{3} \binom{3}{k} b^{3-k} (-2)^k
$$

Now, we will calculate each term in the expansion:

1. For $$ k = 0 $$:
   $$
   \binom{3}{0} b^{3-0} (-2)^0 = 1 \cdot b^3 \cdot 1 = b^3
   $$

2. For $$ k = 1 $$:
   $$
   \binom{3}{1} b^{3-1} (-2)^1 = 3 \cdot b^2 \cdot (-2) = -6b^2
   $$

3. For $$ k = 2 $$:
   $$
   \binom{3}{2} b^{3-2} (-2)^2 = 3 \cdot b^1 \cdot 4 = 12b
   $$

4. For $$ k = 3 $$:
   $$
   \binom{3}{3} b^{3-3} (-2)^3 = 1 \cdot 1 \cdot (-8) = -8
   $$

Now, we can combine all these terms together:

$$
(b - 2)^3 = b^3 - 6b^2 + 12b - 8
$$

Thus, the expansion of $$ (\mathrm{b}-2)^{3} $$ is:

$$
b^3 - 6b^2 + 12b - 8
$$

Looking at the options provided, the correct answer is:

$$
\boxed{b^3 - 6b^2 + 12b - 8}
$$
The expansion of $$ (\mathrm{b}-2)^{3} $$ is $$ b^3 - 6b^2 + 12b - 8 $$.

Use the Binomial Theorem to expand

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