$3^{\frac{1}{5}} - 3^{-\frac{1}{5}} = 2a$のとき、次の値を求めよ。 (1) $\sqrt{1+a^2}$ (2) $(a + \sqrt{1+a^2})^5$

1. 問題の内容

$3^{\frac{1}{5}} - 3^{-\frac{1}{5}} = 2a$のとき、次の値を求めよ。

(1) $\sqrt{1+a^2}$

(2) $(a + \sqrt{1+a^2})^5$

2. 解き方の手順

まず、$a = \frac{1}{2}(3^{\frac{1}{5}} - 3^{-\frac{1}{5}})$である。

(1) $\sqrt{1+a^2}$を求める。

$a^2 = \frac{1}{4}(3^{\frac{1}{5}} - 3^{-\frac{1}{5}})^2 = \frac{1}{4}(3^{\frac{2}{5}} - 2\cdot 3^{\frac{1}{5}}3^{-\frac{1}{5}} + 3^{-\frac{2}{5}}) = \frac{1}{4}(3^{\frac{2}{5}} - 2 + 3^{-\frac{2}{5}})$

$1 + a^2 = 1 + \frac{1}{4}(3^{\frac{2}{5}} - 2 + 3^{-\frac{2}{5}}) = \frac{1}{4}(4 + 3^{\frac{2}{5}} - 2 + 3^{-\frac{2}{5}}) = \frac{1}{4}(3^{\frac{2}{5}} + 2 + 3^{-\frac{2}{5}}) = \frac{1}{4}(3^{\frac{1}{5}} + 3^{-\frac{1}{5}})^2$

$\sqrt{1+a^2} = \sqrt{\frac{1}{4}(3^{\frac{1}{5}} + 3^{-\frac{1}{5}})^2} = \frac{1}{2}(3^{\frac{1}{5}} + 3^{-\frac{1}{5}})$

(2) $(a + \sqrt{1+a^2})^5$を求める。

$a + \sqrt{1+a^2} = \frac{1}{2}(3^{\frac{1}{5}} - 3^{-\frac{1}{5}}) + \frac{1}{2}(3^{\frac{1}{5}} + 3^{-\frac{1}{5}}) = \frac{1}{2}(3^{\frac{1}{5}} - 3^{-\frac{1}{5}} + 3^{\frac{1}{5}} + 3^{-\frac{1}{5}}) = \frac{1}{2}(2 \cdot 3^{\frac{1}{5}}) = 3^{\frac{1}{5}}$

$(a + \sqrt{1+a^2})^5 = (3^{\frac{1}{5}})^5 = 3^{\frac{1}{5} \cdot 5} = 3^1 = 3$

3. 最終的な答え

(1) $\sqrt{1+a^2} = \frac{1}{2}(3^{\frac{1}{5}} + 3^{-\frac{1}{5}})$

(2) $(a + \sqrt{1+a^2})^5 = 3$