$(\sqrt{3} - \sqrt{2} + 1)^3 (\sqrt{3} + \sqrt{2} - 1)^3$ を簡単にせよ。代数学式の計算根号展開因数分解2025/6/151. 問題の内容(3−2+1)3(3+2−1)3(\sqrt{3} - \sqrt{2} + 1)^3 (\sqrt{3} + \sqrt{2} - 1)^3(3−2+1)3(3+2−1)3 を簡単にせよ。2. 解き方の手順まず、与えられた式を次のように変形します。[(3−2+1)(3+2−1)]3[(\sqrt{3} - \sqrt{2} + 1)(\sqrt{3} + \sqrt{2} - 1)]^3[(3−2+1)(3+2−1)]3次に、(3−2+1)(3+2−1)(\sqrt{3} - \sqrt{2} + 1)(\sqrt{3} + \sqrt{2} - 1)(3−2+1)(3+2−1) を計算します。(3−2+1)(3+2−1)=[3+(1−2)][3−(1−2)]=(3)2−(1−2)2(\sqrt{3} - \sqrt{2} + 1)(\sqrt{3} + \sqrt{2} - 1) = [\sqrt{3} + (1 - \sqrt{2})][\sqrt{3} - (1 - \sqrt{2})] = (\sqrt{3})^2 - (1 - \sqrt{2})^2(3−2+1)(3+2−1)=[3+(1−2)][3−(1−2)]=(3)2−(1−2)2=3−(1−22+2)=3−(3−22)=3−3+22=22= 3 - (1 - 2\sqrt{2} + 2) = 3 - (3 - 2\sqrt{2}) = 3 - 3 + 2\sqrt{2} = 2\sqrt{2}=3−(1−22+2)=3−(3−22)=3−3+22=22したがって、与えられた式は次のようになります。(22)3=23(2)3=8⋅22=162(2\sqrt{2})^3 = 2^3 (\sqrt{2})^3 = 8 \cdot 2\sqrt{2} = 16\sqrt{2}(22)3=23(2)3=8⋅22=1623. 最終的な答え16216\sqrt{2}162