整式

3次式の実数の範囲で因数分解

by nomura · 2022年1月26日

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3次式の実数の範囲で因数分解
全て実数の範囲で考える。

(1)

a^{3} \pm b^{3} = (a \pm b) (a^{2} \mp a b + b^{2})

(2)

a^{3} \pm 3 b a^{2} + 3 b^{2} a \pm b^{3} = {(a \pm b)}^{3}

(3)

a^{3} + b^{3} + c^{3} + 3 (a + b) (b + c) (c + a) = {(a + b + c)}^{3}

(4)

a b^{2} \pm a^{2} b + b c^{2} \pm b^{2} c + c a^{2} \pm a c^{2} + a b c \pm a b c = (a \pm b) (b \pm c) (c \pm a)

(5)

a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)

(6)

\sum_{i = 1}^{n} \sum_{j = 1}^{n} \sum_{k = 1}^{n} a_{i} a_{j} a_{k} = {(\sum_{k = 1}^{n} a_{k})}^{3}

(1)

\begin{aligned} a^{3} \pm b^{3} & = a^{3} - (\pm a^{2} b - a b^{2}) + (\pm a^{2} b - a b^{2}) \pm b^{3} \\ = a (a^{2} \mp a b + b^{2}) \pm b (a^{2} \mp a b + b^{2}) \\ = (a \pm b) (a^{2} \mp a b + b^{2}) \end{aligned}

(2)

\begin{aligned} a^{3} \pm 3 b a^{2} + 3 b^{2} a \pm b^{3} & = a^{3} \pm b^{3} \pm 3 a b (a \pm b) \\ = (a \pm b) (a^{2} \mp a b + b^{2}) \pm 3 a b (a \pm b) \\ = (a \pm b) (a^{2} \pm 2 a b + b^{2}) \\ = {(a \pm b)}^{3} \end{aligned}

(3)

\begin{aligned} a^{3} + b^{3} + c^{3} + 3 (a + b) (b + c) (c + a) & = a^{3} + b^{3} + c^{3} + 3 (a b^{2} + a^{2} b + b c^{2} + b^{2} c + c a^{2} + a c^{2} + 2 a b c) \\ = a^{3} + 3 (b + c) a^{2} + 3 (b^{2} + c^{2} + 2 b c) a + b^{3} + c^{3} + 3 b c^{2} + 3 b^{2} c \\ = a^{3} + 3 (b + c) a^{2} + 3 {(b + c)}^{2} a + {(b + c)}^{3} \\ = {(a + b + c)}^{3} \end{aligned}

(4)

\begin{aligned} a b^{2} \pm a^{2} b + b c^{2} \pm b^{2} c + c a^{2} \pm a c^{2} + a b c \pm a b c & = (c \pm b) a^{2} + (b^{2} \pm c^{2} + b c \pm b c) a + b c^{2} \pm b^{2} c \\ = (c \pm b) a^{2} + (b (b \pm c) + c (b \pm c)) a + b c (c \pm b) \\ = (c \pm b) a^{2} + (b + c) (b \pm c) a + b c (c \pm b) \\ = (c \pm b) {a^{2} \pm (b + c) a + b c} \\ = (c \pm b) (a \pm b) (a \pm c) \\ = (a \pm b) (b \pm c) (c \pm a) \end{aligned}

(5)

\begin{aligned} a^{3} + b^{3} + c^{3} - 3 a b c & = {(a + b)}^{3} - 3 a b (a + b) + c^{3} - 3 a b c \\ = {(a + b)}^{3} + c^{3} - 3 a b (a + b + c) \\ = (a + b + c) ({(a + b)}^{2} - (a + b) c + c^{2}) - 3 a b (a + b + c) \\ = (a + b + c) ({(a + b)}^{2} - (a + b) c + c^{2} - 3 a b) \\ = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) \end{aligned}

(6)

\begin{aligned} \sum_{i = 1}^{n} \sum_{j = 1}^{n} \sum_{k = 1}^{n} a_{i} a_{j} a_{k} & = (\sum_{i = 1}^{n} a_{i}) (\sum_{j = 1}^{n} a_{j}) (\sum_{k = 1}^{n} a_{k}) \\ = {(\sum_{k = 1}^{n} a_{k})}^{3} \end{aligned}

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