bạn ơi có thiếu +3 hay j ko

\(a^3+b^3+c^3=3abc\)

\(=> (a^3+b^3) + c^3 - 3abc = 0\)

\(=> (a+b)^3 - 3ab(a+b) + c^3 - 3abc=0\)

\(=> [(a+b)^3+c^3] - 3ab(a+b+c) = 0\)

\(=> (a+b+c)[(a+b)^2-c(a+b)+c^2]-3ab(a+b+c)=0\)

\(=> (a+b+c)(a^2+b^2+c^2+2ab-bc-ca)-3ab(a+b+c)=0\)

\(=> (a+b+c)(a^2+b^2+c^2+2ab-bc-ca-3ab)=0\)

\(=> (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0\)

\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

Ta có: \(a=b=c\Rightarrow\hept{\begin{cases}a^3=abc\\a^3=b^3=c^3\end{cases}}\)

Vì \(a^3=b^3=c^3\Rightarrow a^3+b^3+c^3=3a^3\)

\(\Rightarrow a^3+b^3+c^3=3abc\left(đpcm\right)\)

\(a+b+c=0\)

\(\Leftrightarrow a+b=-c\)

\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3=-c^3\)

\(\Leftrightarrow a^3+3ab\left(a+b\right)+b^3+c^3=0\)

\(\Leftrightarrow a^3-3abc+b^3+c^3=0\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\)

Nếu :  a + b + c = 0 
=> a + b = -c 
=> (a + b)³ = -c³ 
=>a³+b³+c³ =-3ab(a + b)=3abc

Chỉ biết vậy thôi!!!!

a^3+b^3+c^3-3abc=0 

<=>(a+b)^3 -3ab(a+b) +c^3 - 3abc=0 

<=>[(a+b)^3 +c^3] -3ab.(a+b+c)=0 

<=>(a+b+c). [(a+b)^2 -c.(a+b)+c^2] -3ab(a+b+c)=0 

<=>(a+b+c).(a^2+2ab+b^2-ca-cb+c^2-3ab)... 

<=>(a+b+c).(a^2+b^2+c^2-ab-bc-ca)=0 

luôn đúng do a+b+c=0

Vay............................

a/ \(a+b+c=0\)

\(\Leftrightarrow\left(a+b+c\right)^3=0\)

\(\Leftrightarrow\left[\left(a+b\right)+c\right]^3=0\)

\(\Leftrightarrow\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3=0\)

\(\Leftrightarrow a^3+b^3+c^3+3a^2b+3ab^2+3bc^2+3b^2c+3a^2c+3ac^2+6abc=0\)

\(\Leftrightarrow a^3+b^3+c^3+\left(3a^2b+3ab^2+3abc\right)+\left(3bc^2+3b^2c+3abc\right)+\left(3ac^2+3a^2c+3abc\right)-3abc=0\)

\(\Leftrightarrow a^3+b^3+c^3+3abc\left(a+b+c\right)+3bc\left(a+b+c\right)+3ac\left(a+b+c\right)-3abc=0\)

Mà \(a+b+c=0\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow a^3+b^3+c^3=3abc\left(đpcm\right)\)

b/ \(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{matrix}\right.\)

+) Nếu : \(a^2+b^2+c^2-ab-bc-ac=0\)

\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)

Vậy \(a^3+b^3+c^3=3abc\) \(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)