数学 - 代数学 - 整式の計算 - 式の展開、等式の証明

学習環境

代数への出発 (新装版数学入門シリーズ) (松坂和夫(著)、岩波書店)の第2章(整式の計算)、練習問題の問4の解答を求めてみる。

1. $\begin{array}{l} \frac{1}{2} ({(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}) \\ = \frac{1}{2} (a^{2} - 2 a b + b^{2} + b^{2} - 2 b c + c^{2} + c^{2} - 2 c a + a^{2}) \\ = a^{2} + b^{2} + c^{2} - a b - b c - c a \end{array}$
2. $\begin{array}{l} {(a c + b d)}^{2} + {(a d - b c)}^{2} \\ = a^{2} c^{2} + b^{2} d^{2} + a^{2} d^{2} + b^{2} c^{2} \\ = (a^{2} + b^{2}) (c^{2} + d^{2}) \end{array}$
3. $\begin{array}{l} a^{2} (b - c) + b^{2} (c - a) + c^{2} (a - b) \\ = (b - c) a^{2} - (b^{2} - c^{2}) a + b c (b - c) \\ = (a^{2} - (b + c) a + b c) (b - c) \\ = (a - b) (a - c) (b - c) \end{array}$

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