已知非零实数a,b,c满足a+b+c=0,求证(1).a^3+b^3+c^3=3abc
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![已知非零实数a,b,c满足a+b+c=0,求证(1).a^3+b^3+c^3=3abc](/uploads/image/z/5082729-33-9.jpg?t=%E5%B7%B2%E7%9F%A5%E9%9D%9E%E9%9B%B6%E5%AE%9E%E6%95%B0a%2Cb%2Cc%E6%BB%A1%E8%B6%B3a%2Bb%2Bc%3D0%2C%E6%B1%82%E8%AF%81%281%29.a%5E3%2Bb%5E3%2Bc%5E3%3D3abc)
已知非零实数a,b,c满足a+b+c=0,求证(1).a^3+b^3+c^3=3abc
已知非零实数a,b,c满足a+b+c=0,求证(1).a^3+b^3+c^3=3abc
已知非零实数a,b,c满足a+b+c=0,求证(1).a^3+b^3+c^3=3abc
a+b=-c
(a+b)^2=c^2
a^2+b^2+2ab=c^2
a^2+b^2=c^2-2ab
a^3+b^3+c^3
=(a+b)(a^2-ab+b^2)+c^3
=(-c)(a^2-ab+b^2)+c^3
=c(c^2-a^2+ab-b^2)
=c[c^2-(a^2+b^2)+ab]
=c(c^2-c^2+2ab+ab)
=c*3ab
=3abc
a^3+b^3+c^3-3abc
=(a+b)^3+c^3-3a^2b-3ab^2-3abc
=(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)
=(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]
因为a+b+c=0
所以a^3+b^3+c^3-3abc=0
a^3+b^3+c^3=3abc
a^3+b^3+c^3=(a+b)(a^2-ab+b^2)+c^3
=-c(a^2-ab+b^2)+c^3
=-c((a+b)^2-3ab)+c^3
=-c((-c)^2-3ab)+c^3
=-c^3+3abc+c^3
=3abc
a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
因为 a+b+c=0
所以 a^3+b^3+c^3-3abc=0
即: a^3+b^3+c^3=3abc
a+b+c=0
(a+b+c)^3=0
a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+3ac^2+3b^2c+3bc^2+9abc=0
a^3+b^3+c^3+3ab(a+b+c)+3ac(a+b+c)+3bc(a+b+c)-3abc=0
a^3+b^3+c^3-3abc=0
a^3+b^3+c^3=3abc
a^3+b^3+c^3-3abc
=(a+b)^3+c^3-3a^2b-3ab^2-3abc
=(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)
=(a+b+c)[(a+b)^2-(a+b)c+c^2-3ab]
由于a+b+c=0
则a^3+b^3+c^3-3abc=0
a^3+b^3+c^3=3abc