複素数と方程式｜３次方程式の解と係数の関係について

$$\begin{align*} &\text{$3$ 次方程式} \\[ 5pt ] &\quad ax^{\scriptsize{3}}+bx^{\scriptsize{2}}+cx+d=0 \\[ 7pt ] &\text{の $3$ つの解を} \\[ 5pt ] &\quad \alpha \ , \ \beta \ , \ \gamma \\[ 7pt ] &\text{とすると} \\[ 5pt ] &\text{$[1]$ 解と係数の関係} \\[ 5pt ] &\quad \alpha+\beta+\gamma=-\frac{b}{a} \\[ 7pt ] &\quad \alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a} \\[ 7pt ] &\quad \alpha \beta \gamma=-\frac{d}{a} \\[ 7pt ] &\text{$[2]$ 因数分解} \\[ 5pt ] &\quad ax^{\scriptsize{3}}+bx^{\scriptsize{2}}+cx+d=a \left(x-\alpha \right) \left(x-\beta \right) \left(x-\gamma \right) \end{align*}$$

$$\begin{align*} &\text{$3$ 次方程式} \\[ 5pt ] &\quad ax^{\scriptsize{3}}+bx^{\scriptsize{2}}+cx+d=0 \\[ 7pt ] &\text{の $3$ つの解を} \\[ 5pt ] &\quad \alpha \ , \ \beta \ , \ \gamma \\[ 7pt ] &\text{とする。} \\[ 5pt ] &\quad P(x)=ax^{\scriptsize{3}}+bx^{\scriptsize{2}}+cx+d \\[ 7pt ] &\text{とすると、$\alpha \ , \ \beta \ , \ \gamma$ が $P(x)=0$ の $3$ つの解であるので、} \\[ 5pt ] &\text{$k$ を定数とすると、次の等式が成り立つ。} \\[ 5pt ] &\quad ax^{\scriptsize{3}}+bx^{\scriptsize{2}}+cx+d=k \left(x-\alpha \right) \left(x-\beta \right) \left(x-\gamma \right) \\[ 7pt ] &\text{両辺の $x^{\scriptsize{3}}$ の項の係数を比較すると、$k=a$ より} \\[ 5pt ] &\quad ax^{\scriptsize{3}}+bx^{\scriptsize{2}}+cx+d=a\left(x-\alpha \right) \left(x-\beta \right) \left(x-\gamma \right) \\[ 7pt ] &\text{が得られる。この等式の右辺を展開すると} \\[ 5pt ] &\quad \text{(右辺)} \ =ax^{\scriptsize{3}}-a\left(\alpha+\beta+\gamma \right)x^{\scriptsize{2}}+a\left(\alpha \beta+\beta \gamma+\gamma \alpha \right)x-a\alpha \beta \gamma \\[ 7pt ] &\text{等式の両辺を比較すると} \\[ 5pt ] &\quad b=-a\left(\alpha+\beta+\gamma \right) \\[ 7pt ] &\quad c=a\left(\alpha \beta+\beta \gamma+\gamma \alpha \right) \\[ 7pt ] &\quad d=-a\alpha \beta \gamma \\[ 7pt ] &\text{したがって} \\[ 5pt ] &\quad \alpha+\beta+\gamma=-\frac{b}{a} \\[ 7pt ] &\quad \alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a} \\[ 7pt ] &\quad \alpha \beta \gamma=-\frac{d}{a} \end{align*}$$

$$\begin{align*} &\text{$3$ 次方程式} \\[ 5pt ] &\quad x^{\scriptsize{3}}+x^{\scriptsize{2}}+x+3=0 \\[ 7pt ] &\text{の $3$ つの解を} \\[ 5pt ] &\quad \alpha \ , \ \beta \ , \ \gamma \\[ 7pt ] &\text{とするとき、次の式の値を求めよ。} \\[ 5pt ] &(1) \quad {\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}} \\[ 7pt ] &(2) \quad {\alpha}^{\scriptsize{3}}+{\beta}^{\scriptsize{3}}+{\gamma}^{\scriptsize{3}} \end{align*}$$

$$\begin{align*} &\text{$3$ 次方程式} \\[ 5pt ] &\quad x^{\scriptsize{3}}+x^{\scriptsize{2}}+x+3=0 \\[ 7pt ] &\text{の $3$ つの解を} \\[ 5pt ] &\quad \alpha \ , \ \beta \ , \ \gamma \\[ 7pt ] &\text{とするとき、次の式の値を求めよ。} \\[ 5pt ] &\quad {\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}} \end{align*}$$

$$\begin{align*} &\quad x^{\scriptsize{3}}+x^{\scriptsize{2}}+x+3=0 \\[ 7pt ] &\text{解と係数の関係より} \\[ 5pt ] &\quad \alpha+\beta+\gamma=-\frac{1}{1}=-1 \\[ 7pt ] &\quad \alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{1}=1 \\[ 7pt ] &\quad \alpha \beta \gamma=-\frac{3}{1}=-3 \end{align*}$$

$$\begin{align*} &\quad \vdots \\[ 7pt ] &\quad \alpha \beta \gamma=-\frac{3}{1}=-3 \\[ 7pt ] &\text{与式を変形すると} \\[ 5pt ] &\quad {\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}} =\left(\alpha+\beta+\gamma \right)^{\scriptsize{2}}-2\left(\alpha \beta+\beta \gamma+\gamma \alpha \right) \end{align*}$$

$$\begin{align*} &\quad \vdots \\[ 7pt ] &\quad {\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}} =\left(\alpha+\beta+\gamma \right)^{\scriptsize{2}}-2\left(\alpha \beta+\beta \gamma+\gamma \alpha \right) \\[ 7pt ] &\text{よって} \\[ 5pt ] &\quad {\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}}=\left(-1 \right)^{\scriptsize{2}}-2 \cdot 1 \\[ 7pt ] &\quad \therefore \ {\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}}= -1 \end{align*}$$

$$\begin{align*} &\quad \left(\alpha+\beta+\gamma \right)^{\scriptsize{2}}={\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}}+2\alpha \beta+2\beta \gamma+2\gamma \alpha \\[ 7pt ] &\text{より} \\[ 5pt ] &\quad {\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}} =\left(\alpha+\beta+\gamma \right)^{\scriptsize{2}}-2\alpha \beta-2\beta \gamma-2\gamma \alpha \\[ 7pt ] &\text{よって} \\[ 5pt ] &\quad {\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}} =\left(\alpha+\beta+\gamma \right)^{\scriptsize{2}}-2\left(\alpha \beta+\beta \gamma+\gamma \alpha \right) \end{align*}$$

$$\begin{align*} &\text{$3$ 次方程式} \\[ 5pt ] &\quad x^{\scriptsize{3}}+x^{\scriptsize{2}}+x+3=0 \\[ 7pt ] &\text{の $3$ つの解を} \\[ 5pt ] &\quad \alpha \ , \ \beta \ , \ \gamma \\[ 7pt ] &\text{とするとき、次の式の値を求めよ。} \\[ 5pt ] &\quad {\alpha}^{\scriptsize{3}}+{\beta}^{\scriptsize{3}}+{\gamma}^{\scriptsize{3}} \end{align*}$$

$$\begin{align*} &\quad x^{\scriptsize{3}}+x^{\scriptsize{2}}+x+3=0 \\[ 7pt ] &\text{解と係数の関係より} \\[ 5pt ] &\quad \alpha+\beta+\gamma=-\frac{1}{1}=-1 \\[ 7pt ] &\quad \alpha \beta+\beta \gamma+\gamma \alpha=\frac{1}{1}=1 \\[ 7pt ] &\quad \alpha \beta \gamma=-\frac{3}{1}=-3 \end{align*}$$

$$\begin{align*} &\quad \vdots \\[ 7pt ] &\quad \alpha \beta \gamma=-\frac{3}{1}=-3 \\[ 7pt ] &\text{与式を変形すると} \\[ 5pt ] &\quad {\alpha}^{\scriptsize{3}}+{\beta}^{\scriptsize{3}}+{\gamma}^{\scriptsize{3}} \ =\left(\alpha+\beta+\gamma \right) \left({\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}}-\alpha \beta-\beta \gamma-\gamma \alpha \right)+3\alpha \beta \gamma \end{align*}$$

$$\begin{align*} &\quad \vdots \\[ 7pt ] &\quad {\alpha}^{\scriptsize{3}}+{\beta}^{\scriptsize{3}}+{\gamma}^{\scriptsize{3}} =\left(\alpha+\beta+\gamma \right) \left({\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}}-\alpha \beta-\beta \gamma-\gamma \alpha \right)+3\alpha \beta \gamma \\[ 7pt ] &\text{よって} \\[ 5pt ] &\quad {\alpha}^{\scriptsize{3}}+{\beta}^{\scriptsize{3}}+{\gamma}^{\scriptsize{3}} =\left(-1 \right) \left(-1-1 \right)+3 \cdot \left(-3 \right) \\[ 7pt ] &\quad \therefore \ {\alpha}^{\scriptsize{3}}+{\beta}^{\scriptsize{3}}+{\gamma}^{\scriptsize{3}} =-7 \end{align*}$$

$$\begin{align*} &{\alpha}^{\scriptsize{3}}+{\beta}^{\scriptsize{3}}+{\gamma}^{\scriptsize{3}}-3\alpha \beta \gamma \\[ 7pt ] = &\left(\alpha+\beta \right)^{\scriptsize{3}}-3\alpha \beta \left(\alpha+\beta \right)+{\gamma}^{\scriptsize{3}}-3\alpha \beta \gamma \\[ 7pt ] = &\left(\alpha+\beta \right)^{\scriptsize{3}}+{\gamma}^{\scriptsize{3}}-3\alpha \beta \left(\alpha+\beta+\gamma \right) \\[ 7pt ] = &\left\{ \left(\alpha+\beta \right)+{\gamma} \right\} \left\{ \left(\alpha+\beta \right)^{\scriptsize{2}}-\gamma \left(\alpha+\beta \right)+{\gamma}^{\scriptsize{2}} \right\}-3\alpha \beta \left(\alpha+\beta+\gamma \right) \\[ 7pt ] = &\left(\alpha+\beta+\gamma \right) \left\{ \left(\alpha+\beta \right)^{\scriptsize{2}}-\gamma \left(\alpha+\beta \right)+{\gamma}^{\scriptsize{2}} \right\}-3\alpha \beta \left(\alpha+\beta+\gamma \right) \\[ 7pt ] = &\left(\alpha+\beta+\gamma \right) \left\{ \left(\alpha+\beta \right)^{\scriptsize{2}}-\gamma \left(\alpha+\beta \right)+{\gamma}^{\scriptsize{2}}-3\alpha \beta \right\} \\[ 7pt ] = &\left(\alpha+\beta+\gamma \right) \left({\alpha}^{\scriptsize{2}}+2\alpha \beta+{\beta}^{\scriptsize{2}}-\gamma \alpha-\beta \gamma+{\gamma}^{\scriptsize{2}}-3\alpha \beta \right) \\[ 7pt ] = &\left(\alpha+\beta+\gamma \right) \left({\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}}-\alpha \beta-\beta \gamma-\gamma \alpha \right) \\[ 7pt ] &\text{よって} \\[ 5pt ] &{\alpha}^{\scriptsize{3}}+{\beta}^{\scriptsize{3}}+{\gamma}^{\scriptsize{3}} =\left(\alpha+\beta+\gamma \right) \left({\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}}-\alpha \beta-\beta \gamma-\gamma \alpha \right)+3\alpha \beta \gamma \end{align*}$$

$$\begin{align*} &{\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}} =\left(\alpha+\beta+\gamma \right)^{\scriptsize{2}}-2\left(\alpha \beta+\beta \gamma+\gamma \alpha \right) \\[ 7pt ] &{\alpha}^{\scriptsize{3}}+{\beta}^{\scriptsize{3}}+{\gamma}^{\scriptsize{3}} =\left(\alpha+\beta+\gamma \right) \left({\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}}-\alpha \beta-\beta \gamma-\gamma \alpha \right)+3\alpha \beta \gamma \end{align*}$$

$$\begin{align*} &\text{$\alpha \ , \ \beta \ , \ \gamma$ は $3$ 次方程式の解であるので、} \\[ 5pt ] &\quad {\alpha}^{\scriptsize{3}}+{\alpha}^{\scriptsize{2}}+{\alpha}+3=0 \\[ 7pt ] &\quad {\beta}^{\scriptsize{3}}+{\beta}^{\scriptsize{2}}+{\beta}+3=0 \\[ 7pt ] &\quad {\gamma}^{\scriptsize{3}}+{\gamma}^{\scriptsize{2}}+{\gamma}+3=0 \\[ 7pt ] &\text{よって} \\[ 5pt ] &\quad {\alpha}^{\scriptsize{3}}=-{\alpha}^{\scriptsize{2}}-{\alpha}-3 \\[ 7pt ] &\quad {\beta}^{\scriptsize{3}}=-{\beta}^{\scriptsize{2}}-{\beta}-3 \\[ 7pt ] &\quad {\gamma}^{\scriptsize{3}}=-{\gamma}^{\scriptsize{2}}-{\gamma}-3 \\[ 7pt ] &\text{これらを与式に代入すると} \\[ 5pt ] &\quad {\alpha}^{\scriptsize{3}}+{\beta}^{\scriptsize{3}}+{\gamma}^{\scriptsize{3}} =\left(-{\alpha}^{\scriptsize{2}}-{\alpha}-3 \right)+\left(-{\beta}^{\scriptsize{2}}-{\beta}-3 \right)+\left(-{\gamma}^{\scriptsize{2}}-{\gamma}-3 \right) \\[ 7pt ] &\text{整理すると} \\[ 5pt ] &\quad {\alpha}^{\scriptsize{3}}+{\beta}^{\scriptsize{3}}+{\gamma}^{\scriptsize{3}} =-\left({\alpha}^{\scriptsize{2}}+{\beta}^{\scriptsize{2}}+{\gamma}^{\scriptsize{2}} \right)-\left({\alpha}+{\beta}+{\gamma} \right)-9 \\[ 7pt ] &\text{よって、(1)の結果と解と係数の関係から} \\[ 5pt ] &\quad {\alpha}^{\scriptsize{3}}+{\beta}^{\scriptsize{3}}+{\gamma}^{\scriptsize{3}} =-\left(-1 \right)-\left(-1 \right)-9 \\[ 7pt ] &\quad \therefore \ {\alpha}^{\scriptsize{3}}+{\beta}^{\scriptsize{3}}+{\gamma}^{\scriptsize{3}} =-7 \end{align*}$$