与えられた4つの式を展開する問題です。代数学展開多項式因数分解2025/7/231. 問題の内容与えられた4つの式を展開する問題です。2. 解き方の手順(1) (a−b−c)2(a-b-c)^2(a−b−c)2(a−b−c)2=(a−b−c)(a−b−c)(a-b-c)^2 = (a-b-c)(a-b-c)(a−b−c)2=(a−b−c)(a−b−c)=a(a−b−c)−b(a−b−c)−c(a−b−c)= a(a-b-c) - b(a-b-c) - c(a-b-c)=a(a−b−c)−b(a−b−c)−c(a−b−c)=a2−ab−ac−ab+b2+bc−ac+bc+c2= a^2 - ab - ac - ab + b^2 + bc - ac + bc + c^2=a2−ab−ac−ab+b2+bc−ac+bc+c2=a2+b2+c2−2ab−2ac+2bc= a^2 + b^2 + c^2 - 2ab - 2ac + 2bc=a2+b2+c2−2ab−2ac+2bc(2) (x+y)(x−y)(x2+y2)(x+y)(x-y)(x^2+y^2)(x+y)(x−y)(x2+y2)(x+y)(x−y)=x2−y2(x+y)(x-y) = x^2 - y^2(x+y)(x−y)=x2−y2(x2−y2)(x2+y2)=(x2)2−(y2)2(x^2-y^2)(x^2+y^2) = (x^2)^2 - (y^2)^2(x2−y2)(x2+y2)=(x2)2−(y2)2=x4−y4= x^4 - y^4=x4−y4(3) (x+3y)2(x−3y)2(x+3y)^2(x-3y)^2(x+3y)2(x−3y)2(x+3y)(x−3y)=x2−9y2(x+3y)(x-3y) = x^2 - 9y^2(x+3y)(x−3y)=x2−9y2(x+3y)2(x−3y)2=[(x+3y)(x−3y)]2(x+3y)^2(x-3y)^2 = [(x+3y)(x-3y)]^2(x+3y)2(x−3y)2=[(x+3y)(x−3y)]2=(x2−9y2)2= (x^2 - 9y^2)^2=(x2−9y2)2=(x2)2−2(x2)(9y2)+(9y2)2= (x^2)^2 - 2(x^2)(9y^2) + (9y^2)^2=(x2)2−2(x2)(9y2)+(9y2)2=x4−18x2y2+81y4= x^4 - 18x^2y^2 + 81y^4=x4−18x2y2+81y4(4) (x2+x−2)(x2−x−2)(x^2+x-2)(x^2-x-2)(x2+x−2)(x2−x−2)(x2−2+x)(x2−2−x)=(x2−2)2−x2(x^2-2+x)(x^2-2-x) = (x^2-2)^2 - x^2(x2−2+x)(x2−2−x)=(x2−2)2−x2=(x2)2−2(x2)(2)+22−x2= (x^2)^2 - 2(x^2)(2) + 2^2 - x^2=(x2)2−2(x2)(2)+22−x2=x4−4x2+4−x2= x^4 - 4x^2 + 4 - x^2=x4−4x2+4−x2=x4−5x2+4= x^4 - 5x^2 + 4=x4−5x2+43. 最終的な答え(1) a2+b2+c2−2ab−2ac+2bca^2 + b^2 + c^2 - 2ab - 2ac + 2bca2+b2+c2−2ab−2ac+2bc(2) x4−y4x^4 - y^4x4−y4(3) x4−18x2y2+81y4x^4 - 18x^2y^2 + 81y^4x4−18x2y2+81y4(4) x4−5x2+4x^4 - 5x^2 + 4x4−5x2+4