3次方程式 $x^3 - 3x^2 - 2x + 7 = 0$ の3つの解を $\alpha, \beta, \gamma$ とするとき、以下の式の値を求めよ。 (1) $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$ (2) $\alpha^2 + \beta^2 + \gamma^2$ (3) $\alpha^3 + \beta^3 + \gamma^3$ (4) $(1-\alpha)(1-\beta)(1-\gamma)$ (5) $(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$

1. 問題の内容

3次方程式 $x^3 - 3x^2 - 2x + 7 = 0$ の3つの解を $\alpha, \beta, \gamma$ とするとき、以下の式の値を求めよ。

(1) $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$

(2) $\alpha^2 + \beta^2 + \gamma^2$

(3) $\alpha^3 + \beta^3 + \gamma^3$

(4) $(1-\alpha)(1-\beta)(1-\gamma)$

(5) $(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$

2. 解き方の手順

3次方程式の解と係数の関係より、

$\alpha + \beta + \gamma = 3$

$\alpha\beta + \beta\gamma + \gamma\alpha = -2$

$\alpha\beta\gamma = -7$

(1)

$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{-2}{-7} = \frac{2}{7}$

(2)

$\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 3^2 - 2(-2) = 9 + 4 = 13$

(3)

$\alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha)$

$\alpha^3 + \beta^3 + \gamma^3 = (\alpha+\beta+\gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha) + 3\alpha\beta\gamma$

$\alpha^3 + \beta^3 + \gamma^3 = (\alpha+\beta+\gamma)[(\alpha+\beta+\gamma)^2 - 3(\alpha\beta+\beta\gamma+\gamma\alpha)] + 3\alpha\beta\gamma$

$\alpha^3 + \beta^3 + \gamma^3 = 3[3^2 - 3(-2)] + 3(-7) = 3(9+6) - 21 = 45 - 21 = 24$

または、$\alpha, \beta, \gamma$ は $x^3 = 3x^2 + 2x - 7$ の解なので、

$\alpha^3 = 3\alpha^2 + 2\alpha - 7$

$\beta^3 = 3\beta^2 + 2\beta - 7$

$\gamma^3 = 3\gamma^2 + 2\gamma - 7$

$\alpha^3 + \beta^3 + \gamma^3 = 3(\alpha^2+\beta^2+\gamma^2) + 2(\alpha+\beta+\gamma) - 21 = 3(13) + 2(3) - 21 = 39 + 6 - 21 = 24$

(4)

$(1-\alpha)(1-\beta)(1-\gamma) = 1 - (\alpha+\beta+\gamma) + (\alpha\beta + \beta\gamma + \gamma\alpha) - \alpha\beta\gamma = 1 - 3 - 2 - (-7) = 1 - 3 - 2 + 7 = 3$

(5)

$(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha) = (3-\gamma)(3-\alpha)(3-\beta) = 27 - 9(\alpha+\beta+\gamma) + 3(\alpha\beta+\beta\gamma+\gamma\alpha) - \alpha\beta\gamma = 27 - 9(3) + 3(-2) - (-7) = 27 - 27 - 6 + 7 = 1$

3. 最終的な答え

(1) $\frac{2}{7}$

(2) $13$

(3) $24$

(4) $3$

(5) $1$