Question

.

102の(2)解き方教えてください！！

102 次の式を因数分解せよ。
(1) 2y2ー3xyー2y2ー10x+5y十12 (2)% x%(ッーる) 十ア"(タータ)十<?(ァーツ)
(3⑬) 223g2)(Z2二8g二一27 (4% 9二%9上%二ァ2二ァ十1
(5 9十2ーァッキダリ十yツ       1(G)計07ニ2CE7E2CVSEiUaat

Answer 1

ひかる

x^3(y-z)+y^3(z-x)+z^3(x-y)
=(y-z)x^3-(y^3-z^3)x+(y^2-z^2)yz
=(y-z)x^3-(y-z)(y^2+yz+z^2)x+(y-z)(y+z)yz
=(y-z){x^3-(y^2+yz+z^2)x+(y+z)yz}
=(y-z)(x^3-xy^2-xyz-xz^2+y^2z+yz^2)
=(y-z){(z-x)y^2+(z-x)yz-(z^2-x^2)x}
=(y-z){(z-x)y^2+(z-x)yz-(z-x)(z+x)x}
=(y-z)(z-x)(y^2+yz-x^2-xz)
=(y-z)(z-x){(y-x)(y+x)+(y-x)z)}
=(y-z)(z-x)(y-x)(y+x+z)
=-(x-y)(y-z)(z-x)(x+y+z)