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\begin{document}
\begin{math}
問~~~F=(x+y+z)^5-x^5-y^5-z^5 ~~ を因数分解せよ。\\[0.2cm]
解~~~F=5(x^4y+x^4z+y^4x+y^4z+z^4x+z^4+y)\\
~~~~~+10(x^3y^2+x^3z^2+y^3x^2+y^3z^2+z^3x^2+z^3y^2)\\
~~~~~+20(x^3yz+y^3xz+z^3xy)\\
~~~~~+30(x^2y^2z+x^2z^2y+y^2z^2x)\\[0.2cm]
Step~1~~~(4,1,0)~~を求める。\\
(x+y+z)^3(xy+xz+yz)\\
=(x^3+y^3+z^3+3x^2y+3x^2z+3y^2x+3y^2z+3z^2x+3z^2y+6xyz)\\
(xy+xz+yz)\\
=x^4y+y^4x+z^3xy+3x^3y^2+3x^3yz+3y^3x^2+3y^3xz+3z^2x^2y+3z^2y^2x+6x^2y^2z\\
+x^4z+y^3xz+z^4x+3x^3yz+3x^3z^2+3y^2x^2z+3y^2z^2x+3z^3x^2+3z^3xy+6x^2z^2y\\
+x^3yz+y^4z+z^4y+3x^2y^2z+3x^2z^2y+3y^3xz+3y^3z^2+3z^3xy+3z^3y^2+6y^2z^2x\\
=(4,1,0)+7(3,1,1)+3(3,2,0)+12(2,2,1)\\
\therefore (4,1,1)=s_1^3s_2-7(3,1,1)-3(3,2,0)-12(2,2,1)\\[0.2cm]
Step~2~~~(3,2,0)~~を求める。\\
(x+y+z)(xy+xz+yz)^2\\
=(x+y+z)(x^2y^2+x^2z^2+y^2z^2+2x^2yz+2y^2xz+2z^2xy)\\
=x^3y^2+x^3z^2+y^2z^2x+2x^3yz+2y^2x^2z+2z^2x^2y\\
+y^3x^2+x^2z^2y+y^3z^2+2x^2y^2z+2y^3xz+2z^2y^2x\\
+x^2y^2z+z^3x^2+z^3y^2+2x^2z^2y+2y^2z^2x+2z^3xy\\
=(3,2,0)+5(2,2,1)+2(3,1,1)\\
\therefore (3,2,0)=s_1s_2^2-5(2,2,1)-2(3,1,1)\\[0.2cm]
Step~3~~~(3,1,1)~~を求める。\\
(x+y+z)^2xyz\\
=(x^2+y^2+z^2+2xy+2xz+2yz)xyz\\
=x^3yz+y^3xz+z^3xy+2x^2y^2z+2x^2z^2y+2y^2z^2x\\
=(3,1,1)+2(2,2,1)\\
\therefore (3,1,1)=s_1^2s_3-2(2,2,1)\\[0.2cm]
Step~4~~~(2,2,1)~~を求める。\\
(xy+xz+yz)xyz=x^2y^2z+x^2z^2y+y^2z^2x=(2,2,1)\\
\therefore (2,2,1)=s_2s_3\\[0.2cm]
\therefore (3,1,1)=s_1^2s_3-2s_2s_3\\[0.2cm]
\therefore (3,2,0)=s_1s_2^2-5(2,2,1)-2(3,1,1)\\
~~~~~~~~~~~~~~=s_1s_2^2-5s_2s_3-2(s_1^2s_3-2s_2s_3)\\
~~~~~~~~~~~~~~=s_1s_2^2-s_2s_3-2s_1^2s_3\\
\therefore (4,1,1)=s_1^3s_2-7(3,1,1)-3(3,2,0)-12(2,2,1)\\
~~~~~~~~~~~~~~=s_1^3s_2-7(s_1^2s_3-2s_2s_3)\\
~~~~~~~~~~~~~~-3(s_1s_2^2-s_2s_3-2s_1^2s_3)\\
~~~~~~~~~~~~~~-12(s_2s_3)\\
~~~~~~~~~~~~~~=s_1^3s_2-s_1^2s_3-3s_1s_2^2+5s_2s_3\\
\therefore 与式/5=(4,1,0)+2(3,2,0)+4(3,1,1)+6(2,2,1)\\
~~~~~~~~~~~~~~=s_1^3s_2-s_1^2s_3-3s_1s_2^2+5s_2s_3\\
~~~~~~~~~~~~~~+2(s_1s_2^2-s_2s_3-2s_1^2s_3)\\
~~~~~~~~~~~~~~+4(s_1^2s_3-2s_2s_3)+6s_2s_3\\
~~~~~~~~~~~~~~=s_1^3s_2-s_1s_2^2-s_1^2s_3+s_2s_3\\
~~~~~~~~~~~~~~=s_1^2(s_1s_2-s_3)-s_2(s_1s_2-s_3)\\
~~~~~~~~~~~~~~=(s_1s_2-s_3)(s_1^2-s_2)\\
~~~~~(s_1s_2-s_3)~を計算する。\\
~~~~~~~~(s_1s_2-s_3)=(x+y+z)(xy+xz+yz)-xyz\\
~~~~~~~~=x^2y+x^2z+xyz+y^2x+xyz+y^2z+xyz+z^2x+z^2y-xyz\\
~~~~~~~~=(y+z)x^2+(y+z)^2x+(y+z)yz\\
~~~~~~~~=(y+z)(x^2+(y+z)x+yz)=(y+z)(x+y)(x+z)\\
~~~~~(s_1^2-s_3)~を計算する。\\
~~~~~~~(s_1^2-s_3)=x^2+y^2+z^2+xy+xz+yz\\
\therefore 与式=5(y+z)(x+y)(x+z)(x^2+y^2+z^2+xy+xz+yz)\\
~~~~~~~~~~~~~~~~~~~~~~~~~~(解おわり)
\end{math}
\end{document}