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 

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)

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

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

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

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

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

Ta có:\(a^3+b^3+c^3-3abc\)

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

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

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

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

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

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

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