$x = \frac{1}{\sqrt{2}+1}$ , $y = \frac{1}{\sqrt{2}-1}$ のとき、以下の式の値を求めよ。 (1) $x^2 + y^2$ (2) $x^3 + y^3$ (3) $x^3 y - x^2 y^2 + x y^3$

1. 問題の内容

$x = \frac{1}{\sqrt{2}+1}$ , $y = \frac{1}{\sqrt{2}-1}$ のとき、以下の式の値を求めよ。

(1) $x^2 + y^2$

(2) $x^3 + y^3$

(3) $x^3 y - x^2 y^2 + x y^3$

2. 解き方の手順

まず、$x$と$y$を簡単にします。

$x = \frac{1}{\sqrt{2}+1} = \frac{\sqrt{2}-1}{(\sqrt{2}+1)(\sqrt{2}-1)} = \frac{\sqrt{2}-1}{2-1} = \sqrt{2}-1$

$y = \frac{1}{\sqrt{2}-1} = \frac{\sqrt{2}+1}{(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{\sqrt{2}+1}{2-1} = \sqrt{2}+1$

次に、$x+y$ と $xy$ の値を求めます。

$x+y = (\sqrt{2}-1) + (\sqrt{2}+1) = 2\sqrt{2}$

$xy = (\sqrt{2}-1)(\sqrt{2}+1) = 2-1 = 1$

(1) $x^2 + y^2$ を計算します。

$x^2 + y^2 = (x+y)^2 - 2xy = (2\sqrt{2})^2 - 2(1) = 8 - 2 = 6$

(2) $x^3 + y^3$ を計算します。

$x^3 + y^3 = (x+y)(x^2 - xy + y^2) = (x+y)((x+y)^2 - 3xy) = (2\sqrt{2})((2\sqrt{2})^2 - 3(1)) = (2\sqrt{2})(8 - 3) = 10\sqrt{2}$

(3) $x^3 y - x^2 y^2 + x y^3$ を計算します。

$x^3 y - x^2 y^2 + x y^3 = xy(x^2 - xy + y^2) = xy((x+y)^2 - 3xy) = 1((2\sqrt{2})^2 - 3(1)) = 1(8 - 3) = 5$

3. 最終的な答え

(1) $x^2 + y^2 = 6$

(2) $x^3 + y^3 = 10\sqrt{2}$

(3) $x^3 y - x^2 y^2 + x y^3 = 5$