Fc Alan Walker
04/12/2017 at 22:00-
KEITA FC 8C 04/12/2017 at 22:10\(a^3+b^3+c-3abc\)
\(=\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc\)
\(=\left[\left(a+b\right)^3+c^3\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ba-ca\right)\)
Fc Alan Walker selected this answer. -
Mons Trần 04/12/2017 at 22:07We have:a3+b3+c3-3abc
=a3+3a2b+3ab2+b3+c3-3abc-3a2b-3ab2
=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)[(a+b)2-(a+b).c+c2]-3ab(a+b+c)
=(a+b+c)(a2+b2+c2-ac-ab-bc)
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Alchemy 05/12/2017 at 11:28KEITA FC 8C: TAO NGHI MAY KO NEN TU HOI TU TRA LOI :(