数学 – 文字と記号の活躍 - 整式の除法と分数式 – 因数分解(商, 剰余, 2種類以上の文字を含む整式)

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数学読本〈1〉数・式の計算/方程式/不等式 (松坂和夫(著)、岩波書店)の第2章(文字と記号の活躍 - 式の計算)、2.3(整式の除法と分数式)、整式の除法の問15.を解いてみる。

問15.

$\begin{array}{l} (a^{2} + a b + b^{2}) a - b a^{2} - b^{2} a - b^{3} \\ = (a^{2} + a b + b^{2}) (a - b) \end{array}$
$\begin{array}{l} (a - b) a^{3} + b a^{3} - b^{4} \\ = (a - b) (a^{3} + b a^{2}) + b^{2} a^{2} - b^{4} \\ = (a - b) (a^{3} + b a^{2} + b^{2} a) - b^{3} a - b^{4} \\ = (a - b) (a^{3} + b a^{2} + b^{2} a - b^{3}) \end{array}$
$\begin{array}{l} (a + (b + c)) a^{2} - (b + c) a^{2} - (3 b c) a + (b^{3} + c^{3}) \\ = (a + (b + c)) (a^{2} - (b + c) a) + ({(b + c)}^{2} - 3 b c) a + (b^{3} + c^{3}) \\ = (a + (b + c)) (a^{2} - (b + c) a + ({(b + c)}^{2} - 3 b c)) + (b^{3} + c^{3}) - (b + c) ({(b + c)}^{2} - 3 b c) \\ = (a + b + c) (a^{2} + b^{2} + c^{2} - b c - c a - a b) + b^{3} + c^{3} - (b + c) (b^{2} + c^{2} - b c) \\ = (a + b + c) (a^{2} + b^{2} + c^{2} - b c - c a - a b) - b c^{2} + b^{2} c - c b^{2} + b c^{2} \\ = (a + b + c) (a^{2} + b^{2} + c^{2} - b c - c a - a b) \end{array}$

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