与えられた3つの式を展開する問題です。 (1) $(a+b+c)^2$ (2) $(a+b+5)^2$ (3) $(a-b-c)^2$代数学展開多項式2025/5/141. 問題の内容与えられた3つの式を展開する問題です。(1) (a+b+c)2(a+b+c)^2(a+b+c)2(2) (a+b+5)2(a+b+5)^2(a+b+5)2(3) (a−b−c)2(a-b-c)^2(a−b−c)22. 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+ba+b2+bc+ca+cb+c2= a^2 + ab + ac + ba + b^2 + bc + ca + cb + c^2=a2+ab+ac+ba+b2+bc+ca+cb+c2=a2+b2+c2+2ab+2bc+2ca= a^2 + b^2 + c^2 + 2ab + 2bc + 2ca=a2+b2+c2+2ab+2bc+2ca(2) (a+b+5)2(a+b+5)^2(a+b+5)2(a+b+5)2=(a+b+5)(a+b+5)(a+b+5)^2 = (a+b+5)(a+b+5)(a+b+5)2=(a+b+5)(a+b+5)=a(a+b+5)+b(a+b+5)+5(a+b+5)= a(a+b+5) + b(a+b+5) + 5(a+b+5)=a(a+b+5)+b(a+b+5)+5(a+b+5)=a2+ab+5a+ba+b2+5b+5a+5b+25= a^2 + ab + 5a + ba + b^2 + 5b + 5a + 5b + 25=a2+ab+5a+ba+b2+5b+5a+5b+25=a2+b2+25+2ab+10a+10b= a^2 + b^2 + 25 + 2ab + 10a + 10b=a2+b2+25+2ab+10a+10b(3) (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−ba+b2+bc−ca+cb+c2= a^2 - ab - ac - ba + b^2 + bc - ca + cb + c^2=a2−ab−ac−ba+b2+bc−ca+cb+c2=a2+b2+c2−2ab+2bc−2ca= a^2 + b^2 + c^2 - 2ab + 2bc - 2ca=a2+b2+c2−2ab+2bc−2ca3. 最終的な答え(1) a2+b2+c2+2ab+2bc+2caa^2 + b^2 + c^2 + 2ab + 2bc + 2caa2+b2+c2+2ab+2bc+2ca(2) a2+b2+2ab+10a+10b+25a^2 + b^2 + 2ab + 10a + 10b + 25a2+b2+2ab+10a+10b+25(3) a2+b2+c2−2ab+2bc−2caa^2 + b^2 + c^2 - 2ab + 2bc - 2caa2+b2+c2−2ab+2bc−2ca