3次式の実数の範囲で因数分解
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。
全て実数の範囲で考える。
(1)
\[ a^{3}\pm b^{3}=\left(a\pm b\right)\left(a^{2}\mp ab+b^{2}\right) \](2)
\[ a^{3}\pm3ba^{2}+3b^{2}a\pm b^{3}=\left(a\pm b\right)^{3} \](3)
\[ a^{3}+b^{3}+c^{3}+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=\left(a+b+c\right)^{3} \](4)
\[ ab^{2}\pm a^{2}b+bc^{2}\pm b^{2}c+ca^{2}\pm ac^{2}+abc\pm abc=\left(a\pm b\right)\left(b\pm c\right)\left(c\pm a\right) \](5)
\[ a^{3}+b^{3}+c^{3}-3abc=\left(a+b+c\right)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right) \](6)
\[ \sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{k=1}^{n}a_{i}a_{j}a_{k}=\left(\sum_{k=1}^{n}a_{k}\right)^{3} \](1)
\begin{align*} a^{3}\pm b^{3} & =a^{3}-\left(\pm a^{2}b-ab^{2}\right)+\left(\pm a^{2}b-ab^{2}\right)\pm b^{3}\\ & =a\left(a^{2}\mp ab+b^{2}\right)\pm b\left(a^{2}\mp ab+b^{2}\right)\\ & =\left(a\pm b\right)\left(a^{2}\mp ab+b^{2}\right) \end{align*}(2)
\begin{align*} a^{3}\pm3ba^{2}+3b^{2}a\pm b^{3} & =a^{3}\pm b^{3}\pm3ab\left(a\pm b\right)\\ & =(a\pm b)\left(a^{2}\mp ab+b^{2}\right)\pm3ab\left(a\pm b\right)\\ & =\left(a\pm b\right)\left(a^{2}\pm2ab+b^{2}\right)\\ & =\left(a\pm b\right)^{3} \end{align*}(3)
\begin{align*} a^{3}+b^{3}+c^{3}+3\left(a+b\right)\left(b+c\right)\left(c+a\right) & =a^{3}+b^{3}+c^{3}+3\left(ab^{2}+a^{2}b+bc^{2}+b^{2}c+ca^{2}+ac^{2}+2abc\right)\\ & =a^{3}+3\left(b+c\right)a^{2}+3\left(b^{2}+c^{2}+2bc\right)a+b^{3}+c^{3}+3bc^{2}+3b^{2}c\\ & =a^{3}+3\left(b+c\right)a^{2}+3\left(b+c\right)^{2}a+\left(b+c\right)^{3}\\ & =\left(a+b+c\right)^{3} \end{align*}(4)
\begin{align*} ab^{2}\pm a^{2}b+bc^{2}\pm b^{2}c+ca^{2}\pm ac^{2}+abc\pm abc & =\left(c\pm b\right)a^{2}+\left(b^{2}\pm c^{2}+bc\pm bc\right)a+bc^{2}\pm b^{2}c\\ & =\left(c\pm b\right)a^{2}+\left(b\left(b\pm c\right)+c\left(b\pm c\right)\right)a+bc\left(c\pm b\right)\\ & =\left(c\pm b\right)a^{2}+\left(b+c\right)\left(b\pm c\right)a+bc\left(c\pm b\right)\\ & =\left(c\pm b\right)\left\{ a^{2}\pm\left(b+c\right)a+bc\right\} \\ & =\left(c\pm b\right)\left(a\pm b\right)\left(a\pm c\right)\\ & =\left(a\pm b\right)\left(b\pm c\right)\left(c\pm a\right) \end{align*}(5)
\begin{align*} a^{3}+b^{3}+c^{3}-3abc & =\left(a+b\right)^{3}-3ab\left(a+b\right)+c^{3}-3abc\\ & =\left(a+b\right)^{3}+c^{3}-3ab\left(a+b+c\right)\\ & =\left(a+b+c\right)\left(\left(a+b\right)^{2}-\left(a+b\right)c+c^{2}\right)-3ab\left(a+b+c\right)\\ & =\left(a+b+c\right)\left(\left(a+b\right)^{2}-\left(a+b\right)c+c^{2}-3ab\right)\\ & =\left(a+b+c\right)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right) \end{align*}(6)
\begin{align*} \sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{k=1}^{n}a_{i}a_{j}a_{k} & =\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{j=1}^{n}a_{j}\right)\left(\sum_{k=1}^{n}a_{k}\right)\\ & =\left(\sum_{k=1}^{n}a_{k}\right)^{3} \end{align*}ページ情報
| タイトル | 3次式の実数の範囲で因数分解 |
| URL | https://www.nomuramath.com/zp9df08z/ |
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複二次式の定義と因数分解
\[
a_{4}x^{4}+a_{2}x^{2}+a_{0}=\frac{1}{4a_{4}}\left(2a_{4}x^{2}+a_{2}+\sqrt{a_{2}^{\;2}-4a_{4}a_{0}}\right)\left(2a_{4}x^{2}+a_{2}-\sqrt{a_{2}^{\;2}-4a_{4}a_{0}}\right)
\]
n乗根の因数分解
\[
z^{n}-1=\prod_{k=1}^{n}\left(z-e^{\frac{2\pi}{n}ki}\right)
\]
ビネ・コーシーとラグランジュの恒等式
\[
\left(\sum_{i=1}^{n}a_{i}c_{i}\right)\left(\sum_{j=1}^{n}b_{j}d_{j}\right)-\left(\sum_{i=1}^{n}a_{i}d_{i}\right)\left(\sum_{j=1}^{n}b_{j}c_{j}\right)=\sum_{1\leq i<j\leq n}\left(a_{i}b_{j}-a_{j}b_{i}\right)\left(c_{i}d_{j}-c_{j}d_{i}\right)
\]
因数分解による3次方程式の標準形の解
\[
x_{k}=\omega^{k}\sqrt[3]{-\frac{q}{2}+\sqrt{\left(\frac{q}{2}\right)^{2}+\left(\frac{p}{3}\right)^{3}}}-\omega^{3-k}\frac{p}{3}\frac{1}{\sqrt[3]{-\frac{q}{2}-\sqrt{\left(\frac{q}{2}\right)^{2}+\left(\frac{p}{3}\right)^{3}}}}\cnd{k\in\left\{ 0,1,2\right\} }
\]