次の3つの式を計算する問題です。 (1) $\sqrt[3]{54} - \sqrt[3]{250} + \sqrt[3]{16}$ (2) $\sqrt{2} \div \sqrt[4]{4} \times \sqrt[12]{32} \div \sqrt[6]{2}$ (3) $(\sqrt[3]{16} + \sqrt[3]{8} + \sqrt[3]{4})(\sqrt[3]{4} - \sqrt[3]{2})$

1. 問題の内容

次の3つの式を計算する問題です。

(1) $\sqrt[3]{54} - \sqrt[3]{250} + \sqrt[3]{16}$

(2) $\sqrt{2} \div \sqrt[4]{4} \times \sqrt[12]{32} \div \sqrt[6]{2}$

(3) $(\sqrt[3]{16} + \sqrt[3]{8} + \sqrt[3]{4})(\sqrt[3]{4} - \sqrt[3]{2})$

2. 解き方の手順

(1)

$\sqrt[3]{54} = \sqrt[3]{27 \cdot 2} = 3\sqrt[3]{2}$

$\sqrt[3]{250} = \sqrt[3]{125 \cdot 2} = 5\sqrt[3]{2}$

$\sqrt[3]{16} = \sqrt[3]{8 \cdot 2} = 2\sqrt[3]{2}$

よって、

$\sqrt[3]{54} - \sqrt[3]{250} + \sqrt[3]{16} = 3\sqrt[3]{2} - 5\sqrt[3]{2} + 2\sqrt[3]{2} = (3 - 5 + 2)\sqrt[3]{2} = 0\sqrt[3]{2} = 0$

(2)

$\sqrt[4]{4} = (4)^{\frac{1}{4}} = (2^2)^{\frac{1}{4}} = 2^{\frac{2}{4}} = 2^{\frac{1}{2}} = \sqrt{2}$

$\sqrt[12]{32} = (32)^{\frac{1}{12}} = (2^5)^{\frac{1}{12}} = 2^{\frac{5}{12}}$

$\sqrt[6]{2} = 2^{\frac{1}{6}}$

よって、

$\sqrt{2} \div \sqrt[4]{4} \times \sqrt[12]{32} \div \sqrt[6]{2} = \sqrt{2} \div \sqrt{2} \times 2^{\frac{5}{12}} \div 2^{\frac{1}{6}} = 1 \times 2^{\frac{5}{12} - \frac{1}{6}} = 2^{\frac{5}{12} - \frac{2}{12}} = 2^{\frac{3}{12}} = 2^{\frac{1}{4}} = \sqrt[4]{2}$

(3)

$\sqrt[3]{16} = \sqrt[3]{2^4} = 2\sqrt[3]{2}$

$\sqrt[3]{8} = 2$

$\sqrt[3]{4} = \sqrt[3]{2^2}$

$(\sqrt[3]{16} + \sqrt[3]{8} + \sqrt[3]{4})(\sqrt[3]{4} - \sqrt[3]{2}) = (2\sqrt[3]{2} + 2 + \sqrt[3]{4})(\sqrt[3]{4} - \sqrt[3]{2})$

$= 2\sqrt[3]{2} \cdot \sqrt[3]{4} - 2\sqrt[3]{2} \cdot \sqrt[3]{2} + 2\sqrt[3]{4} - 2\sqrt[3]{2} + \sqrt[3]{4} \cdot \sqrt[3]{4} - \sqrt[3]{4} \cdot \sqrt[3]{2}$

$= 2\sqrt[3]{8} - 2\sqrt[3]{4} + 2\sqrt[3]{4} - 2\sqrt[3]{2} + \sqrt[3]{16} - \sqrt[3]{8}$

$= 2 \cdot 2 - 2\sqrt[3]{4} + 2\sqrt[3]{4} - 2\sqrt[3]{2} + 2\sqrt[3]{2} - 2$

$= 4 - 2 = 2$

3. 最終的な答え

(1) 0

(2) $\sqrt[4]{2}$

(3) 2