ここで勉強すれば数学検定１級の壁は超えられるか。

 

次の式を係数が整数の範囲で因数分解する。

 

[math]1-x^{2}-y^{2}-z^{2}+2xyz-\left( x-yz\right) \left( y-zx\right) -\left( y-zx\right) \left( z-xy\right)-\left( z-xy\right) \left( x-yz\right)[/math]

 

 

 

（ｘ＝１を上の代入して０になることを確認。（ｘ－１）の因数があると予想。同様にｙ＝１、ｚ＝１を上の式に代入して０になることを確認。すなわち、（ｘ－１）（ｙ－１）（ｚ－１）（・・・）の形になることを予想して因数分解する。）

 

 

 

 

 

 

式を展開して、ｘの項べきの順に並べる。

 

[math]1-x^{2}-y^{2}-z^{2}+2xyz-\left( x-yz\right) \left( y-zx\right) -\left( y-zx\right) \left( z-xy\right)-\left( z-xy\right) \left( x-yz\right)[/math]

 

 

[math]=1-x^{2}-y^{2}-z^{2}+2xyz-\left( xy-x^{2}z-y^{2}z+xyz^{2}\right)-\left( yz-xy^{2}-xz^{2}+x^{2}yz\right) -\left( xz-yz^{2}-x^{2}y+xy^{2}z\right)[/math]

 

 

[math]=x^{2}\left( -yz+y+z-1\right) +x\left( -y^{2}z+y^{2}-yz^{2}+2yz-y+z^{2}-z\right)+y^{2}z-y^{2}+yz^{2}-yz-z^{2}+1[/math]

 

 

[math]=x^{2}\left\{ -y\left( z-1\right) +\left( z-1\right) \right\} +x\left\{ \left( y+z\right) ^2-yz\left( y+z\right) -\left( y+z\right) \right\}+y^{2}\left( z-1\right) +yz\left( z-1\right) -\left( z^{2}-1\right)[/math]

 

 

 

[math]=x^{2}\left( z-1\right) \left( -y+1\right) +x\left( y+z\right) \left\{ \left( y+z\right) -yz-1\right\} +\left( z-1\right) \left\{ y^{2}+yz-\left( z+1\right) \right\}[/math]より

 

 

 

[math]=x^{2}\left( z-1\right) \left( -y+1\right) +x\left( y+z\right) \left\{ -y\left( z-1\right) +\left( z-1\right) \right\}+\left( z-1\right) \left\{ z\left( y-1\right) +y^{2}-1\right\}[/math]

 

 

 

[math]=-x^{2}\left( x-1\right) \left( y-1\right) +x\left( y+z\right) \left( z-1\right) \left( -y+1\right)+\left( z-1\right) \left( y-1\right) \left\{ z+\left( y+1\right) \right\}[/math]

 

 

 

[math]=-x^{2}\left( z-1\right) \left( y-1\right) -x\left( y+z\right) \left( z-1\right) \left( y-1\right) +\left( z-1\right) \left( y-1\right) \left( y+z+1\right)[/math]

 

 

 

[math]=-\left( z-1\right) \left( y-1\right) \left\{ x^{2}+x\left( y+z\right) -\left( y+z+1\right) \right\}[/math]

 

 

 

[math]=-\left( z-1\right) \left( y-1\right) \left\{ x+\left( y+z+1\right) \right\} \left( x+1\right)[/math]

 

 

 

[math]=-\left( x-1\right) \left( y-1\right) \left( z-1\right) \left( x+y+z+1\right)[/math]　

 

 

 

 

[math]-\left( x-1\right) \left( y-1\right) \left( z-1\right) \left( x+y+z+1\right)[/math]・・・答え