| ■25209 / inTopicNo.4) |
Re[3]: お願いします
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□投稿者/ X 大御所(264回)-(2007/05/26(Sat) 17:41:11)
 | (x+y+z)^3-x^3-y^3-z^3
={(x+y+z)^3-x^3}-{y^3+z^3}
=(y+z){(x+y+z)^2+(x+y+z)x+x^2}-(y+z)(y^2-yz+z^2)
=(y+z){(x+y+z)^2+(x+y+z)x+x^2-(y^2-yz+z^2)}
=3(y+z)(x^2+xy+zx+yz)
=3(y+z){x^2+(y+z)x+yz}
=3(x+y)(y+z)(z+x)
となります。
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