\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}

To expand \( (\mathrm{b}-2)^{3} \) using the Binomial Theorem, we start by identifying the coefficients based on the theorem's formula. The general form for the expansion of \( (x + y)^n \) is \( \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \). In your case, \( x = b \), \( y = -2 \), and \( n = 3 \). Using the coefficients, we get: \[ (\mathrm{b} - 2)^3 = \sum_{k=0}^{3} \binom{3}{k} \mathrm{b}^{3-k} (-2)^k \] Calculating each term: - For \( k = 0 \): \( \binom{3}{0} b^3 (-2)^0 = 1 \cdot b^3 \cdot 1 = b^3 \) - For \( k = 1 \): \( \binom{3}{1} b^2 (-2)^1 = 3 \cdot b^2 \cdot (-2) = -6b^2 \) - For \( k = 2 \): \( \binom{3}{2} b^1 (-2)^2 = 3 \cdot b \cdot 4 = 12b \) - For \( k = 3 \): \( \binom{3}{3} b^0 (-2)^3 = 1 \cdot 1 \cdot (-8) = -8 \) So, combining all the terms, we have: \[ b^3 - 6b^2 + 12b - 8 \] Thus, the expanded form of \( (\mathrm{b}-2)^{3} \) is \( b^{3}-6 b^{2}+12 b-8 \). Now, to find the correct choice from the given options, the correct expansion is: \[ b^{3}-6 b^{2}+12 b-8 \] So, the correct choice is: \( b^{3}-6 b^{2}+12 b-8 \)