$x = 3 + 2\sqrt{2}$、$y = 3 - 2\sqrt{2}$ のとき、以下の式の値を求めよ。 (1) $x+y$ (2) $xy$ (3) $x^2+y^2$ (4) $x^3+y^3$ (5) $x^5+y^5$

1. 問題の内容

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

(1) $x+y$

(2) $xy$

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

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

(5) $x^5+y^5$

2. 解き方の手順

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

(2) $xy = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1$

(3) $x^2 + y^2 = (x+y)^2 - 2xy = 6^2 - 2(1) = 36 - 2 = 34$

または、

$x^2 + y^2 = (3+2\sqrt{2})^2 + (3-2\sqrt{2})^2 = (9 + 12\sqrt{2} + 8) + (9 - 12\sqrt{2} + 8) = 17 + 17 = 34$

(4) $x^3 + y^3 = (x+y)(x^2 - xy + y^2) = (x+y)((x^2 + y^2) - xy) = 6(34 - 1) = 6(33) = 198$

または、

$x^3 + y^3 = (3+2\sqrt{2})^3 + (3-2\sqrt{2})^3 = (27 + 54\sqrt{2} + 72 + 16\sqrt{2}) + (27 - 54\sqrt{2} + 72 - 16\sqrt{2}) = 27 + 72 + 27 + 72 = 198$

(5) $x^5 + y^5 = (x^2 + y^2)(x^3 + y^3) - x^2y^3 - x^3y^2 = (x^2 + y^2)(x^3 + y^3) - x^2y^2(x + y) = (34)(198) - (1)^2(6) = 6732 - 6 = 6726$

3. 最終的な答え

(1) $x+y = 6$

(2) $xy = 1$

(3) $x^2+y^2 = 34$

(4) $x^3+y^3 = 198$

(5) $x^5+y^5 = 6726$