Mathematics
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(5) (a+b+c)(a+b2+c-ab-bc-ca)
={a+(b+c)}{a-(b+c)a+b2-bc+c2}
=a-(b+c)a²+(b2-bc+c2)a
+(b+c)a²-(b+c)2a+(b+c)(b2-bc+c2)
1つずつ項をかけ算
すると式が長くなる
SU
ので,a以外の文字
は定数と考える
='+{(62-bc+c2)-(62+2bc+c2)}a+6-b2c+bc+b2c-bc+c
=a3-3abc+b3+ c³=a+b+c³-3abc
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大変助かりました!
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