n乗の3つの数の和を基本対称式で表す

n乗の3つの数の和を基本対称式で表す

$\alpha ,\beta ,\gamma$を解とする3次方程式を考える． $$\begin{array}{rcl}(x-\alpha )(x-\beta )(x-\gamma )& =& 0\\ x^3-(\alpha +\beta +\gamma )x^2+(\alpha \beta +\beta \gamma +\alpha \gamma )x-\alpha \beta \gamma & =& 0\end{array}$$ $\alpha ,\beta ,\gamma$は式の解なので，$x$に代入しても式は成り立つ．代入した2つの式を得る． $$\begin{array}{r}\{\begin{array}{lll}{\alpha }^3-(\alpha +\beta +\gamma ){\alpha }^2+(\alpha \beta +\beta \gamma +\alpha \gamma )\alpha -\alpha \beta \gamma & =& 0\\ {\beta }^3-(\alpha +\beta +\gamma ){\beta }^2+(\alpha \beta +\beta \gamma +\alpha \gamma )\beta -\alpha \beta \gamma & =& 0\\ {\gamma }^3-(\alpha +\beta +\gamma ){\gamma }^2+(\alpha \beta +\beta \gamma +\alpha \gamma )\gamma -\alpha \beta \gamma & =& 0\end{array}\end{array}$$ 第一式には${\alpha }^{n-3}$，第二式には${\beta }^{n-3}$，第三式には${\gamma }^{n-3}$を両辺に掛ける． $$\begin{array}{r}\{\begin{array}{lll}{\alpha }^{n-3}\{{\alpha }^3-(\alpha +\beta +\gamma ){\alpha }^2+(\alpha \beta +\beta \gamma +\alpha \gamma )\alpha -\alpha \beta \gamma \}& =& {\alpha }^{n-3}\cdot 0\\ {\beta }^{n-3}\{{\beta }^3-(\alpha +\beta +\gamma ){\beta }^2+(\alpha \beta +\beta \gamma +\alpha \gamma )\beta -\alpha \beta \gamma \}& =& {\beta }^{n-3}\cdot 0\\ {\gamma }^{n-3}\{{\gamma }^3-(\alpha +\beta +\gamma ){\gamma }^2+(\alpha \beta +\beta \gamma +\alpha \gamma )\gamma -\alpha \beta \gamma \}& =& {\gamma }^{n-3}\cdot 0\end{array}\\ \\ \{\begin{array}{llllll}{\alpha }^n& -(\alpha +\beta +\gamma ){\alpha }^{n-1}& +(\alpha \beta +\beta \gamma +\alpha \gamma ){\alpha }^{n-2}& -\alpha \beta \gamma {\alpha }^{n-3}& =& 0\\ {\beta }^n& -(\alpha +\beta +\gamma ){\beta }^{n-1}& +(\alpha \beta +\beta \gamma +\alpha \gamma ){\beta }^{n-2}& -\alpha \beta \gamma {\beta }^{n-3}& =& 0\\ {\gamma }^n& -(\alpha +\beta +\gamma ){\gamma }^{n-1}& +(\alpha \beta +\beta \gamma +\alpha \gamma ){\gamma }^{n-2}& -\alpha \beta \gamma {\gamma }^{n-3}& =& 0\end{array}\end{array}$$ 3式を足し合わせ，n乗の和について解くことで．n乗の和を基本対称式で求める式を得る． $$\begin{array}{rclrclr}& {\alpha }^n& -(\alpha +\beta +\gamma ){\alpha }^{n-1}& +(\alpha \beta +\beta \gamma +\alpha \gamma ){\alpha }^{n-2}& -\alpha \beta \gamma {\alpha }^{n-3}& =& 0\\ & {\beta }^n& -(\alpha +\beta +\gamma ){\beta }^{n-1}& +(\alpha \beta +\beta \gamma +\alpha \gamma ){\beta }^{n-2}& -\alpha \beta \gamma {\beta }^{n-3}& =& 0\\ +)& {\gamma }^n& -(\alpha +\beta +\gamma ){\gamma }^{n-1}& +(\alpha \beta +\beta \gamma +\alpha \gamma ){\gamma }^{n-2}& -\alpha \beta \gamma {\gamma }^{n-3}& =& 0\\ & {\alpha }^n+{\beta }^n+{\gamma }^n& -(\alpha +\beta +\gamma )({\alpha }^{n-1}+{\beta }^{n-1}+{\gamma }^{n-1})& +(\alpha \beta +\beta \gamma +\alpha \gamma )({\alpha }^{n-2}+{\beta }^{n-2}+{\gamma }^{n-2})& -\alpha \beta \gamma ({\alpha }^{n-3}+{\beta }^{n-3}+{\gamma }^{n-3})& =& 0\end{array}$$ $$\begin{array}{rcl}{\alpha }^n+{\beta }^n+{\gamma }^n& =& (\alpha +\beta +\gamma )({\alpha }^{n-1}+{\beta }^{n-1}+{\gamma }^{n-1})-(\alpha \beta +\beta \gamma +\alpha \gamma )({\alpha }^{n-2}+{\beta }^{n-2}+{\gamma }^{n-2})+\alpha \beta \gamma ({\alpha }^{n-3}+{\beta }^{n-3}+{\gamma }^{n-3})\end{array}$$

---

$\alpha ,\beta ,\gamma$は任意の数でよいのでそれを$x,y,z$とすれば以下の式を得る． $$\begin{array}{rcl}{x}^n+y^n+z^n& =& (x+y+z)(x^{n-1}+y^{n-1}+z^{n-1})-(xy+yz+xz)(x^{n-2}+y^{n-2}+z^{n-2})+xyz(x^{n-3}+y^{n-3}+z^{n-3})\end{array}$$