Ta có :

Giả thuyết : a + b + c = 0

(a + b + c)³ = 0

a³ + b³ + c³ + 3.(a + b)(b + c)(c + a) = 0

Từ a + b + c = 0

=> \(\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)

=> a³ + b³ + c³ + 3.(-c)(-a)(-b) = 0

=> a³ + b³ + c³ = 3abc 

a+b+c=0
a+b=-c
(a+b)^3=(-c)^3
a^3+3a^2b+3ab^2+b^3=(-c)^3
a^3+b^3+c^3=-3a^2b-3ab^2
a^3+b^3+c^3=-3ab(-c)
a^3+b^3+c^3=3abc

Ta có : \(a+b+c=0\Rightarrow-a-b=c\)

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

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

\(\Rightarrow-3a^2b-3ab^2-3abc=3ab\left(-a-b\right)-3abc\)

\(=3abc-3abc=0\) (đpccm)

\(\frac{a^3+b^3+c^3-3abc}{a+b+c}=\frac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{a+b+c}=\frac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a+b+c}\)

\(=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{a+b+c}=a^2+b^2+c^2-ab-bc-ca\)

\(=\frac{1}{2}\left(a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\right)\)

\(=\frac{1}{2}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\ge0\) (đpcm)

 thay a^3+b^3=(a+b)^3 -3ab(a+b) .Ta có : 

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

a³ + b³ + c³ = 3abc

⇒ a³ + b³ + c³ - 3abc = 0

⇒ ( a³ + b³ ) + c³ - 3abc = 0

⇒ ( a + b )³ - 3ab( a + b ) + c³ - 3abc = 0

⇒ [ ( a + b )³ + c³ ] - [ 3ab( a + b ) + 3abc ] = 0

⇒ ( a + b + c )[ ( a + b )² - ( a + b ).c + c² ] - 3ab( a + b + c ) = 0

⇒ ( a + b + c )( a² + b² + c² - ab - bc - ac ) = 0

⇒ \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)

+) a² + b² + c² - ab - bc - ac = 0

⇒ 2( a² + b² + c² - ab - bc - ac ) = 2.0

⇒ 2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0

⇒ ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( a² - 2ac + c² ) = 0

⇒ ( a - b )² + ( b - c )² + ( a - c )² = 0

VT ≥ 0 ∀ a,b,c . Dấu "=" xảy ra khi a = b = c

⇒ a + b + c = 0 hoặc a = b = c ( đpcm )

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

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

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

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

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

Ta có:\(a^3+b^3+c^3-3abc\)

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

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

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)

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

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