$x = \frac{\sqrt{3}+1}{\sqrt{3}-1}$、 $y = \frac{\sqrt{3}-1}{\sqrt{3}+1}$ のとき、次の式の値を求めます。 (1) $x^3y + xy^3$ (2) $4x^2 + 7xy + 4y^2$ (3) $x^3 + y^3$

1. 問題の内容

$x = \frac{\sqrt{3}+1}{\sqrt{3}-1}$、 $y = \frac{\sqrt{3}-1}{\sqrt{3}+1}$ のとき、次の式の値を求めます。

(1) $x^3y + xy^3$

(2) $4x^2 + 7xy + 4y^2$

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

2. 解き方の手順

まず、$x$ と $y$ を簡単にします。

$x = \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$

$y = \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$

次に、$x+y$ と $xy$ を求めます。

$x+y = (2+\sqrt{3}) + (2-\sqrt{3}) = 4$

$xy = (2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1$

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

$x^3y + xy^3 = 1(4^2 - 2(1)) = 16 - 2 = 14$

(2) $4x^2 + 7xy + 4y^2 = 4(x^2 + y^2) + 7xy = 4((x+y)^2 - 2xy) + 7xy = 4(x+y)^2 - 8xy + 7xy = 4(x+y)^2 - xy$

$4x^2 + 7xy + 4y^2 = 4(4^2) - 1 = 4(16) - 1 = 64 - 1 = 63$

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

$x^3 + y^3 = (4)(4^2 - 3(1)) = 4(16 - 3) = 4(13) = 52$

3. 最終的な答え

(1) $14$

(2) $63$

(3) $52$