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Ta có \(a^3+b^3+c^3-3abc=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)=0
=> a^3+b^3+c^3=3abc(ĐPCM )
^.^
Gt<=>(a+b+c)^3=0
<=>(a+b)^3+c+3+3(a+b)c(a+b+c)=0
<=>a^3+b^3+c^3+3ab(a+b)=0 (vì a+b+c=0)
<=>a^3+b^3+c^3=3abc (vì a+b=-c)
Ta có: a + b + c = 0
<=> a + b = -c
<=> (a+b)3 = -c3
<=> a3 + b3 + 3ab (a+b) = -c3
<=> a3 + b3 - 3abc = -c3 ( Vì a+b = -c )
<=> a3 + b3 + c2 = 3abc
cách khác:
\(a+b+c=0\)
<=> \(a+b=-c\)
<=> \(\left(a+b\right)^3=-c^3\)
Ta có:
\(a^3+b^3+c^3=a^3+b^3-\left(a+b\right)^3=a^3+b^3-a^3-3ab\left(a+b\right)-b^3\)
\(=-3ab\left(a+b\right)=-3ab.\left(-c\right)=3abc\) (đpcm)
Xin lỗi mink chỉ chọn đc 1 câu trả lời thôi vậy nên mink sẽ tặng các bạn còn lại câu khác .
Từ a+b+c=0 a+b=-c a+c=-b b+c=-a
(a+b+c)
=0
=0
=3
=3
=3
=3
c3+b3+a3=3abc (dpcm)
\(Gt\Leftrightarrow\left(a+b+c\right)^3=0\)
\(\Leftrightarrow\left(a+b\right)^3+c+3+3\left(a+b\right)c\left(a+b+c\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3+3ab\left(a+b\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3=3abc\left(DPCM\right)\)
Code : Breacker
\(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^3=0\)
\(\Rightarrow a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3a^2c+3ac^2+6abc=0\)
\(\Rightarrow a^3+b^3+c^3+\left(3a^2b+3abc+3ab^2\right)+\left(3b^2c+3abc+3bc^2\right)+\left(3a^2c+3abc+3ac^2\right)-3abc=0\)
\(\Rightarrow a^3+b^3+c^3+3ab\left(a+b+c\right)+3bc\left(a+b+c\right)+3ac\left(a+b+c\right)=3abc\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
=> ĐPCM
MMS_Hồ Khánh Châu: hơi dài.
Ta có: \(a+b+c=0\)
\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^3=0\)
\(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(a^3+b^3+c^3-3abc=0\)
\(a^3+b^3+c^3=3abc\)
đpcm
