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

1. 問題の内容

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

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

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

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

2. 解き方の手順

まず、$x$と$y$をそれぞれ有理化する。

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

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

(1) $x^2 + y^2$ を求める。

$x^2 = (\sqrt{2}-1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2}$

$y^2 = (\sqrt{2}+1)^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2}$

$x^2 + y^2 = (3 - 2\sqrt{2}) + (3 + 2\sqrt{2}) = 6$

(2) $x^3 + y^3$ を求める。

$x^3 = (\sqrt{2}-1)^3 = (\sqrt{2})^3 - 3(\sqrt{2})^2(1) + 3(\sqrt{2})(1)^2 - 1^3 = 2\sqrt{2} - 6 + 3\sqrt{2} - 1 = 5\sqrt{2} - 7$

$y^3 = (\sqrt{2}+1)^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(1) + 3(\sqrt{2})(1)^2 + 1^3 = 2\sqrt{2} + 6 + 3\sqrt{2} + 1 = 5\sqrt{2} + 7$

$x^3 + y^3 = (5\sqrt{2} - 7) + (5\sqrt{2} + 7) = 10\sqrt{2}$

(3) $x^3y - x^2y^2 + xy^3$ を求める。

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

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

$x^2 + y^2 = 6$ (既出)

$xy = 1$

$x^2 - xy + y^2 = 6 - 1 = 5$

$x^3y - x^2y^2 + xy^3 = 1(5) = 5$

3. 最終的な答え

(1) 6

(2) $10\sqrt{2}$

(3) 5