uitwerking
$
\begin{array}{l}
\left( {a + b} \right)^n = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}c}
n\\
k\\
\end{array}} \right)} \cdot a^{n - k} \cdot b^k\\
(x - 2)^3 = \left( {\begin{array}{*{20}c}
3\\
0\\
\end{array}} \right) \cdot x^3 \cdot \left( { - 2} \right)^0 + \left( {\begin{array}{*{20}c}
3\\
1\\
\end{array}} \right) \cdot x^2 \cdot \left( { - 2} \right)^1 + \left( {\begin{array}{*{20}c}
3\\
2\\
\end{array}} \right) \cdot x^1 \cdot \left( { - 2} \right)^2 + \left( {\begin{array}{*{20}c}
3\\
3\\
\end{array}} \right) \cdot x^0 \cdot \left( { - 2} \right)^3\\
(x - 2)^3 = \left( {\begin{array}{*{20}c}
3\\
0\\
\end{array}} \right) \cdot x^3 + \left( {\begin{array}{*{20}c}
3\\
1\\
\end{array}} \right) \cdot x^2 \cdot \left( { - 2} \right) + \left( {\begin{array}{*{20}c}
3\\
2\\
\end{array}} \right) \cdot x \cdot \left( { - 2} \right)^2 + \left( {\begin{array}{*{20}c}
3\\
3\\
\end{array}} \right) \cdot \left( { - 2} \right)^3\\
(x - 2)^3 = 1 \cdot x^3 + 3 \cdot x^2 \cdot \left( { - 2} \right) + 3 \cdot x \cdot \left( { - 2} \right)^2 + 1 \cdot \left( { - 2} \right)^3\\
(x - 2)^3 = x^3 - 6x^2 + 12x - 8 \\
\end{array}
$