問題は、式 $(x-y)(x+y)^2 + (y-z)(y+z)^2 + (z-x)(z+x)^2$ を解くことです。代数学因数分解多項式2025/5/151. 問題の内容問題は、式 (x−y)(x+y)2+(y−z)(y+z)2+(z−x)(z+x)2(x-y)(x+y)^2 + (y-z)(y+z)^2 + (z-x)(z+x)^2(x−y)(x+y)2+(y−z)(y+z)2+(z−x)(z+x)2 を解くことです。2. 解き方の手順まず、各項を展開します。(x−y)(x+y)2=(x−y)(x2+2xy+y2)=x3+2x2y+xy2−x2y−2xy2−y3=x3+x2y−xy2−y3(x-y)(x+y)^2 = (x-y)(x^2 + 2xy + y^2) = x^3 + 2x^2y + xy^2 - x^2y - 2xy^2 - y^3 = x^3 + x^2y - xy^2 - y^3(x−y)(x+y)2=(x−y)(x2+2xy+y2)=x3+2x2y+xy2−x2y−2xy2−y3=x3+x2y−xy2−y3(y−z)(y+z)2=(y−z)(y2+2yz+z2)=y3+2y2z+yz2−y2z−2yz2−z3=y3+y2z−yz2−z3(y-z)(y+z)^2 = (y-z)(y^2 + 2yz + z^2) = y^3 + 2y^2z + yz^2 - y^2z - 2yz^2 - z^3 = y^3 + y^2z - yz^2 - z^3(y−z)(y+z)2=(y−z)(y2+2yz+z2)=y3+2y2z+yz2−y2z−2yz2−z3=y3+y2z−yz2−z3(z−x)(z+x)2=(z−x)(z2+2zx+x2)=z3+2z2x+zx2−z2x−2zx2−x3=z3+z2x−zx2−x3(z-x)(z+x)^2 = (z-x)(z^2 + 2zx + x^2) = z^3 + 2z^2x + zx^2 - z^2x - 2zx^2 - x^3 = z^3 + z^2x - zx^2 - x^3(z−x)(z+x)2=(z−x)(z2+2zx+x2)=z3+2z2x+zx2−z2x−2zx2−x3=z3+z2x−zx2−x3次に、これらの結果を足し合わせます。(x3+x2y−xy2−y3)+(y3+y2z−yz2−z3)+(z3+z2x−zx2−x3)=x3−x3+y3−y3+z3−z3+x2y−xy2+y2z−yz2+z2x−zx2=x2y−xy2+y2z−yz2+z2x−zx2(x^3 + x^2y - xy^2 - y^3) + (y^3 + y^2z - yz^2 - z^3) + (z^3 + z^2x - zx^2 - x^3) = x^3 - x^3 + y^3 - y^3 + z^3 - z^3 + x^2y - xy^2 + y^2z - yz^2 + z^2x - zx^2 = x^2y - xy^2 + y^2z - yz^2 + z^2x - zx^2(x3+x2y−xy2−y3)+(y3+y2z−yz2−z3)+(z3+z2x−zx2−x3)=x3−x3+y3−y3+z3−z3+x2y−xy2+y2z−yz2+z2x−zx2=x2y−xy2+y2z−yz2+z2x−zx2この式は、さらに因数分解することができます。x2y−xy2+y2z−yz2+z2x−zx2=x2(y−z)+y2(z−x)+z2(x−y)x^2y - xy^2 + y^2z - yz^2 + z^2x - zx^2 = x^2(y-z) + y^2(z-x) + z^2(x-y)x2y−xy2+y2z−yz2+z2x−zx2=x2(y−z)+y2(z−x)+z2(x−y)=x2(y−z)+y2z−y2x+z2x−z2y= x^2(y-z) + y^2z - y^2x + z^2x - z^2y=x2(y−z)+y2z−y2x+z2x−z2y=x2(y−z)+y2z−y2x+z2x−z2y=x2(y−z)−x(y2−z2)+yz(y−z)= x^2(y-z) + y^2z - y^2x + z^2x - z^2y = x^2(y-z) - x(y^2 - z^2) + yz(y-z)=x2(y−z)+y2z−y2x+z2x−z2y=x2(y−z)−x(y2−z2)+yz(y−z)=x2(y−z)−x(y−z)(y+z)+yz(y−z)= x^2(y-z) - x(y-z)(y+z) + yz(y-z)=x2(y−z)−x(y−z)(y+z)+yz(y−z)=(y−z)[x2−x(y+z)+yz]= (y-z)[x^2 - x(y+z) + yz]=(y−z)[x2−x(y+z)+yz]=(y−z)[x2−xy−xz+yz]= (y-z)[x^2 - xy - xz + yz]=(y−z)[x2−xy−xz+yz]=(y−z)[x(x−y)−z(x−y)]= (y-z)[x(x-y) - z(x-y)]=(y−z)[x(x−y)−z(x−y)]=(y−z)(x−y)(x−z)= (y-z)(x-y)(x-z)=(y−z)(x−y)(x−z)=−(x−y)(y−z)(z−x)= -(x-y)(y-z)(z-x)=−(x−y)(y−z)(z−x)3. 最終的な答え−(x−y)(y−z)(z−x) -(x-y)(y-z)(z-x)−(x−y)(y−z)(z−x)または(x−y)(z−y)(z−x)(x-y)(z-y)(z-x)(x−y)(z−y)(z−x)