$(4a - 3b)^3$ を展開しなさい。代数学展開二項定理多項式2025/5/151. 問題の内容(4a−3b)3(4a - 3b)^3(4a−3b)3 を展開しなさい。2. 解き方の手順(a−b)3=a3−3a2b+3ab2−b3(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3(a−b)3=a3−3a2b+3ab2−b3 の公式を利用します。ここで aaa を 4a4a4a, bbb を 3b3b3b と置き換えます。すると、(4a−3b)3=(4a)3−3(4a)2(3b)+3(4a)(3b)2−(3b)3(4a - 3b)^3 = (4a)^3 - 3(4a)^2(3b) + 3(4a)(3b)^2 - (3b)^3(4a−3b)3=(4a)3−3(4a)2(3b)+3(4a)(3b)2−(3b)3となります。次に、各項を計算します。(4a)3=43a3=64a3(4a)^3 = 4^3 a^3 = 64a^3(4a)3=43a3=64a33(4a)2(3b)=3(16a2)(3b)=3⋅16⋅3a2b=144a2b3(4a)^2(3b) = 3(16a^2)(3b) = 3 \cdot 16 \cdot 3 a^2b = 144a^2b3(4a)2(3b)=3(16a2)(3b)=3⋅16⋅3a2b=144a2b3(4a)(3b)2=3(4a)(9b2)=3⋅4⋅9ab2=108ab23(4a)(3b)^2 = 3(4a)(9b^2) = 3 \cdot 4 \cdot 9 ab^2 = 108ab^23(4a)(3b)2=3(4a)(9b2)=3⋅4⋅9ab2=108ab2(3b)3=33b3=27b3(3b)^3 = 3^3 b^3 = 27b^3(3b)3=33b3=27b3したがって、(4a−3b)3=64a3−144a2b+108ab2−27b3(4a - 3b)^3 = 64a^3 - 144a^2b + 108ab^2 - 27b^3(4a−3b)3=64a3−144a2b+108ab2−27b33. 最終的な答え64a3−144a2b+108ab2−27b364a^3 - 144a^2b + 108ab^2 - 27b^364a3−144a2b+108ab2−27b3