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． $$\begin{eqnarray} \left(x-\alpha\right)\left(x-\beta\right)\left(x-\gamma\right)&=&0 \\x^3-\left(\alpha+\beta+\gamma\right)x^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)x-\alpha\beta\gamma&=&0 \end{eqnarray}$$ \(\alpha,\beta,\gamma\)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． $$ \begin{eqnarray} \left\{ \begin{array}{l} \alpha^3-\left(\alpha+\beta+\gamma\right)\alpha^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\alpha-\alpha\beta\gamma&=&0 \\\beta^3-\left(\alpha+\beta+\gamma\right)\beta^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\beta-\alpha\beta\gamma&=&0 \\\gamma^3-\left(\alpha+\beta+\gamma\right)\gamma^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\gamma-\alpha\beta\gamma&=&0 \end{array} \right. \end{eqnarray} $$ 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． $$ \begin{eqnarray} \left\{ \begin{array}{l} \alpha^{n-3}\left\{\alpha^3-\left(\alpha+\beta+\gamma\right)\alpha^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\alpha-\alpha\beta\gamma\right\}&=&\alpha^{n-3}\cdot0 \\\beta^{n-3}\left\{\beta^3-\left(\alpha+\beta+\gamma\right)\beta^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\beta-\alpha\beta\gamma\right\}&=&\beta^{n-3}\cdot0 \\\gamma^{n-3}\left\{\gamma^3-\left(\alpha+\beta+\gamma\right)\gamma^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\gamma-\alpha\beta\gamma\right\}&=&\gamma^{n-3}\cdot0 \end{array} \right. \\ \\\left\{ \begin{array}{l} \alpha^n&-\left(\alpha+\beta+\gamma\right)\alpha^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\alpha^{n-2}&-\alpha\beta\gamma\alpha^{n-3}&=&0 \\\beta^n&-\left(\alpha+\beta+\gamma\right)\beta^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\beta^{n-2}&-\alpha\beta\gamma\beta^{n-3}&=&0 \\\gamma^n&-\left(\alpha+\beta+\gamma\right)\gamma^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\gamma^{n-2}&-\alpha\beta\gamma\gamma^{n-3}&=&0 \end{array} \right. \end{eqnarray} $$ 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． $$\begin{eqnarray} &\alpha^n&-\left(\alpha+\beta+\gamma\right)\alpha^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\alpha^{n-2}&-\alpha\beta\gamma\alpha^{n-3}&=&0 \\&\beta^n&-\left(\alpha+\beta+\gamma\right)\beta^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\beta^{n-2}&-\alpha\beta\gamma\beta^{n-3}&=&0 \\+)&\gamma^n&-\left(\alpha+\beta+\gamma\right)\gamma^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\gamma^{n-2}&-\alpha\beta\gamma\gamma^{n-3}&=&0 \\\hline &\alpha^{n}+\beta^{n}+\gamma^{n} &-\left(\alpha+\beta+\gamma\right)\left(\alpha^{n-1}+\beta^{n-1}+\gamma^{n-1}\right) &+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\left(\alpha^{n-2}+\beta^{n-2}+\gamma^{n-2}\right) &-\alpha\beta\gamma\left(\alpha^{n-3}+\beta^{n-3}+\gamma^{n-3}\right)&=&0 \end{eqnarray}$$ $$\begin{eqnarray} \alpha^{n}+\beta^{n}+\gamma^{n}&=& \left(\alpha+\beta+\gamma\right)\left(\alpha^{n-1}+\beta^{n-1}+\gamma^{n-1}\right) -\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\left(\alpha^{n-2}+\beta^{n-2}+\gamma^{n-2}\right) +\alpha\beta\gamma\left(\alpha^{n-3}+\beta^{n-3}+\gamma^{n-3}\right) \end{eqnarray}$$

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\(\alpha, \beta, \gamma\)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． $$\begin{eqnarray} x^{n}+y^{n}+z^{n}&=&\left(x+y+z\right)\left(x^{n-1}+y^{n-1}+z^{n-1}\right)-\left(xy+yz+xz\right)\left(x^{n-2}+y^{n-2}+z^{n-2}\right)+xyz\left(x^{n-3}+y^{n-3}+z^{n-3}\right) \end{eqnarray}$$