与えられた式 $xy^2 + yz^2 + zx^2 - x^2y - y^2z - z^2x$ を因数分解する。代数学因数分解多項式2025/6/31. 問題の内容与えられた式 xy2+yz2+zx2−x2y−y2z−z2xxy^2 + yz^2 + zx^2 - x^2y - y^2z - z^2xxy2+yz2+zx2−x2y−y2z−z2x を因数分解する。2. 解き方の手順与えられた式を xxx について整理する。xy2+yz2+zx2−x2y−y2z−z2xxy^2 + yz^2 + zx^2 - x^2y - y^2z - z^2xxy2+yz2+zx2−x2y−y2z−z2x=(z−y)x2+(y2−z2)x+(yz2−y2z)= (z-y)x^2 + (y^2 - z^2)x + (yz^2 - y^2z)=(z−y)x2+(y2−z2)x+(yz2−y2z)=(z−y)x2+(y−z)(y+z)x+yz(z−y)= (z-y)x^2 + (y-z)(y+z)x + yz(z-y)=(z−y)x2+(y−z)(y+z)x+yz(z−y)=(z−y)x2−(z−y)(y+z)x+yz(z−y)= (z-y)x^2 - (z-y)(y+z)x + yz(z-y)=(z−y)x2−(z−y)(y+z)x+yz(z−y)=(z−y)[x2−(y+z)x+yz]= (z-y)[x^2 - (y+z)x + yz]=(z−y)[x2−(y+z)x+yz]=(z−y)(x−y)(x−z)= (z-y)(x-y)(x-z)=(z−y)(x−y)(x−z)=−(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)