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} $