第7章 7.32

$x^3 + px^2 + qx + r = 0$

３つの解が $\alpha,\beta,\gamma$ なので上式の因数分解は$(x-\alpha)(x-\beta)(x-\gamma)=0$となる

$(x-\alpha)(x-\beta)(x-\gamma) = x^3 - (\alpha+\beta+\gamma)x^2 + (\alpha\beta+\beta\gamma+\gamma\alpha)x - \alpha\beta\gamma = 0$

２つの式を比較すると

$-(\alpha+\beta+\gamma) = p \qquad (\alpha\beta+\beta\gamma+\gamma\alpha) = q \qquad -\alpha\beta\gamma = r \qquad \text{となる}$

(1)

$\alpha\beta$、$\beta\gamma$、$\gamma\alpha$が解なので

$(x-\alpha\beta)(x-\beta\gamma)(x-\gamma\alpha) = 0$

$x^3 - (\alpha\beta+\beta\gamma+\gamma\alpha)x^2 + (\alpha\beta^2\gamma+\alpha\beta\gamma^2+\alpha^2\beta\gamma)x - \alpha^2\beta^2\gamma^2 = 0$

$x^3 - (\alpha\beta+\beta\gamma+\gamma\alpha)x^2 + \alpha\beta\gamma(\alpha+\beta+\gamma)x - \alpha^2\beta^2\gamma^2 = 0$

$x^3 - qx^2 + prx - r^2 = 0$

(2)

$\alpha^2$、$\beta^2$、$\gamma^2$が解なので

$(x-\alpha^2)(x-\beta^2)(x-\gamma^2) = 0$

$x^3 - (\alpha^2+\beta^2+\gamma^2)x^2 + (\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2)x - \alpha^2\beta^2\gamma^2 = 0$

$x^3 - \{(\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha)\}x^2 + \{(\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2(\alpha\beta^2\gamma+\alpha\beta\gamma^2+\alpha^2\beta\gamma)\}x - \alpha^2\beta^2\gamma^2 = 0$

$x^3 + (2q-p^2)x^2 + (q^2-2pr)x - r^2 = 0$