与えられた式 $(x+y)(x-2y)(x-y)(x+2y)$ を展開せよ。代数学展開多項式因数分解2025/5/141. 問題の内容与えられた式 (x+y)(x−2y)(x−y)(x+2y)(x+y)(x-2y)(x-y)(x+2y)(x+y)(x−2y)(x−y)(x+2y) を展開せよ。2. 解き方の手順まず、(x+y)(x+y)(x+y) と (x−y)(x-y)(x−y)、(x−2y)(x-2y)(x−2y) と (x+2y)(x+2y)(x+2y) をそれぞれ展開します。(x+y)(x−y)=x2−y2(x+y)(x-y) = x^2 - y^2(x+y)(x−y)=x2−y2(x−2y)(x+2y)=x2−(2y)2=x2−4y2(x-2y)(x+2y) = x^2 - (2y)^2 = x^2 - 4y^2(x−2y)(x+2y)=x2−(2y)2=x2−4y2次に、得られた2つの式を掛け合わせます。(x2−y2)(x2−4y2)(x^2 - y^2)(x^2 - 4y^2)(x2−y2)(x2−4y2)展開します。x2(x2−4y2)−y2(x2−4y2)=x4−4x2y2−x2y2+4y4x^2(x^2 - 4y^2) - y^2(x^2 - 4y^2) = x^4 - 4x^2y^2 - x^2y^2 + 4y^4x2(x2−4y2)−y2(x2−4y2)=x4−4x2y2−x2y2+4y4同類項をまとめます。x4−5x2y2+4y4x^4 - 5x^2y^2 + 4y^4x4−5x2y2+4y43. 最終的な答えx4−5x2y2+4y4x^4 - 5x^2y^2 + 4y^4x4−5x2y2+4y4