$(\sqrt{2} + 1)^3$を計算する問題です。代数学展開二項定理根号計算2025/7/171. 問題の内容(2+1)3(\sqrt{2} + 1)^3(2+1)3を計算する問題です。2. 解き方の手順(2+1)3(\sqrt{2} + 1)^3(2+1)3を二項定理または展開の公式を用いて計算します。(2+1)3=(2)3+3(2)2(1)+3(2)(1)2+(1)3(\sqrt{2} + 1)^3 = (\sqrt{2})^3 + 3(\sqrt{2})^2(1) + 3(\sqrt{2})(1)^2 + (1)^3(2+1)3=(2)3+3(2)2(1)+3(2)(1)2+(1)3=22+3(2)+32+1= 2\sqrt{2} + 3(2) + 3\sqrt{2} + 1=22+3(2)+32+1=22+6+32+1= 2\sqrt{2} + 6 + 3\sqrt{2} + 1=22+6+32+1=52+7= 5\sqrt{2} + 7=52+7または、(2+1)2=(2)2+2(2)(1)+12=2+22+1=3+22(\sqrt{2} + 1)^2 = (\sqrt{2})^2 + 2(\sqrt{2})(1) + 1^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2}(2+1)2=(2)2+2(2)(1)+12=2+22+1=3+22(2+1)3=(2+1)(2+1)2=(2+1)(3+22)(\sqrt{2} + 1)^3 = (\sqrt{2} + 1)(\sqrt{2} + 1)^2 = (\sqrt{2} + 1)(3 + 2\sqrt{2})(2+1)3=(2+1)(2+1)2=(2+1)(3+22)=2(3)+2(22)+1(3)+1(22)= \sqrt{2}(3) + \sqrt{2}(2\sqrt{2}) + 1(3) + 1(2\sqrt{2})=2(3)+2(22)+1(3)+1(22)=32+4+3+22= 3\sqrt{2} + 4 + 3 + 2\sqrt{2}=32+4+3+22=52+7= 5\sqrt{2} + 7=52+73. 最終的な答え52+75\sqrt{2} + 752+7