Mathematics
Senior High
Resolved
この因数分解の+3abcが分かりません。
-3abcではないんですか??
1枚目の結果を利用して因数分解をする問題です。
(1) a³ + b³ +c-3abc
=(a+b)³-3ab(a+b)+c³-3abc
= {(a+b)³ + c³)-3ab(a+b)-3abc
(a+b+c){(a+b)²-(a+b)c+c²}
=
(-3ab(a+b+c
=(a+b+c){(a+b)²-(a+b)c+c²-3ab}
08 = (a+b+c)(a²+2ab+b²-ac-bc+c²-3ab)
=(a+b+c)(a²+ b²+c²-ab-bc-ca)
(2) (x-1)³+(2x-1)³-(3x-2)³
=(x-1)³+(2x-1)³+(-3x+2)³3
より, x-1=α, 2x-1=6, -3x+2=c とおくと.
a+b+c=(x-1)+(2x-1)+(-3x+2)=0
だから,
(x-1)³+(2x-1)³-(3x-2)³
=a³ + b³ + c³
=(a+b+c)(a²+ b²+c²-ab-bc-ca)+3abc
=3(x-1)(2x-1)(-3x+2)
=-3(x-1)(2x-1)(3x-2)
100=3abc
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