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\begin{math}
問~~~F=x^4+y^4+z^4-2x^2y^2-2x^2z^2-y^2z^2~~ を因数分解せよ。\\[0.2cm]
解~~~F=(x^2+y^2-z^2)^2-4x^2y^2~~に気がつけばなんでもないが、\\
~~~~~それに気がつかない時は、次のように基本対称式の定理を使って腕力でも解ける。\\
~~~~F=(x+y+z)^4-4(3,1,0)-6(2,2,0)-12(2,1,1)-2(2,2,0)\\
~~~~~=(x+y+z)^4-4(3,1,0)-12(2,1,1)-8(2,2,0)\\
Step~1~~~(3,1,0)~~を求める。\\
(x+y+z)^2(xy+xz+yz)\\
=(x^2+y^2+z^2+2xy+2xz+2yz)(xy+xz+yz)\\
=x^3y+y^3x+z^2xy+2x^2y^2+2x^2z+2y^2xz\\
+x^2z+y^2xz+z^3x+2x^2yz+2x^2z^2+2z^2xy\\
+x^2yz+y^3z+z^3y+2y^2xz+2z^2xy+2y^2z^2\\
=(3,1,0)+5(2,1,1)+2(2,2,0)\\
\therefore (3,1,1)=s_1^2s_2-5(2,1,1)-2(2,2,0)\\[0.2cm]
Step~2~~~(2,2,0)~~を求める。\\
(xy+xz+yz)^2\\
=(x^2y^2+x^2z^2+y^2z^2+2x^2yz+2y^2xz+2z^2xy)\\
=(2,2,0)+2(2,1,1)\\
\therefore (2,2,0)=s_2^2-2(2,2,1)\\[0.2cm]
Step~3~~~(2,1,1)~~を求める。\\
(x+y+z)xyz\\
=x^2yz+y^2xz+z^2=(2,1,1)\\
\therefore (2,1,1)=s_1s_3\\[0.2cm]
\therefore (2,2,0)=s_2^2-2(2,2,1)\\
~~~~~~~~~~~~~~=s_2^2-2s_1s_3\\
\therefore (3,1,0)=s_1^2s_2-5s_1s_3-2(s_2^2-2s_1s_3)\\
~~~~~~~~~~~~~~=s_1^2s_2-s_1s_3-2s_2^2\\
\therefore 与式=(x+y+z)^4-4(3,1,0)-12(2,1,1)-8(2,2,0)\\
~~~~~~~~~~~~~~=s_1^4-4(s_1^2s_2-s_1s_3-2s_2^2)\\
~~~~~~~~~~~~~~-12(s_1s_3)-8(s_2^2-2s_1s_3)\\
~~~~~~~~~~~~~~=s_1^4+8s_1s_3-4s_1^2s_2\\
~~~~~~~~~~~~~~=s_1(s_1^3-8s_3-4s_1s_2)\\
~~~~~(s_1^3-8s_3-4s_1s_2)~を計算する。\\
~~~~~~~~(s_1^3-8s_3-4s_1s_2)=(x+y+z)^3-8xyz-4(x+y+z)(xy+xz+yz)\\
~~~~~~~~=x^3+y^3+z^3+3(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y)+6xyz+8xyz\\
~~~~~~~~~-4(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+3xyz)\\
~~~~~~~~=x^3+y^3+z^3-(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y)+2xyz\\
~~~~~これはz~で整頓すれば、（何で整頓しても同様に出来るが）\\
~~~~~~~~z^3-(x+y)z^2-(x-y)^2z+(x-y)^2(x+y)\\
~~~~~~~=z^2(z-x-y)-(x-y)^2(z-x-y)\\
~~~~~~~=(z-x-y)(z+x-y)(z-x+y)\\
\therefore 与式=(x+y+z)(z-x-y)(z+x-y)(z-x+y)\\
~~~~~~~~~~~~=(x+y+z)(x+y-z)(x-y+z)(x-y-z)\\
~~~~~~~~~~~~~~~~~~~~~~~~~~(解おわり)
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